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If one end of a focal chord AB of the parabola $y^{2}=8x$ is at $A\left ( \frac{1}{2},-2 \right ),$ then the equation of the tangent to it at B is :
Option: 1 $x+2y+8=0$
Option: 2 $2x-y-24=0$
Option: 3 $x-2y+8=0$
Option: 4 $2x+y-24=0$

D

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Posted by

Shabareesh

The length of the minor axis (along y-axis) of an ellipse in the standard form is $\frac{4}{\sqrt{3}}.$ If this ellipse touches the line, $x+6y=8;$ then its eccentricity is :
Option: 1 $\frac{1}{2}\sqrt{\frac{5}{3}}$

Option: 2 $\frac{1}{2}\sqrt{\frac{11}{3}}$

Option: 3 $\sqrt{\frac{5}{6}}$

Option: 4 $\frac{1}{3}\sqrt{\frac{11}{3}}$

What is Ellipse? -

Ellipse

Standard Equation of Ellipse:

The standard form of the equation of an ellipse with center (0, 0) and major axis on the x-axis is

$\mathbf{\frac{\mathbf{x}^{2}}{\mathbf{a}^{2}}+\frac{\mathbf{y}^{2}}{\mathbf{b}^{2}}=1} \quad \text { where }, \mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)_{(\mathrm{a}>\mathrm{b})}$

a > b
the length of the major axis is 2a
the coordinates of the vertices are (±a, 0)
the length of the minor axis is 2b
the coordinates of the co-vertices are (0, ±b)

-

Equation of Tangent of Ellipse in Parametric Form and Slope Form -

Slope Form:

$\\ {\text { The equation of tangent of slope m to the ellipse, } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { are }} \\ {y=m x \pm \sqrt{a^{2} m^{2}+b^{2}} \text { and coordinate of point of contact is }} \\ {\left(\mp \frac{a^{2} m}{\sqrt{a^{2} m^{2}+b^{2}}}, \pm \frac{b^{2}}{\sqrt{a^{2} m^{2}+b^{2}}}\right)}$

-

$\\\begin{array}{l}{2 \mathrm{b}=\frac{4}{\sqrt{3}} \quad \Rightarrow \quad \mathrm{b}=\frac{2}{\sqrt{3}}} \\ {\text { Equation of tangent } \equiv \mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}}\end{array}\\\text { comparing with } \equiv y=\frac{-x}{6}+\frac{4}{3}\\$

$\\ {\mathrm{m}=\frac{-1}{6} \text { and } \mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}=\frac{16}{9}} \\ {\Rightarrow \quad \frac{\mathrm{a}^{2}}{36}+\frac{4}{3}=\frac{16}{9}} \\ {\Rightarrow \quad \frac{\mathrm{a}^{2}}{36}=\frac{16}{9}-\frac{4}{3}=\frac{4}{9}} \\ {\Rightarrow \quad a^{2}=16} \\ {\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}}\\e=\sqrt{\frac{11}{12}}$

Correct Option (2)

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Posted by

avinash.dongre

If $e_{1}\: \: and\: \: e_{2}$ are the eccentricities of the ellipse, $\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$ and the hyperbola, $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ respectively and $\left ( e_{1},e_{2} \right )$ is a point on the ellipse, $15x^{2}+3y^{2}=k,$ then $k$ is equal to :
Option: 1 $14$
Option: 2 $15$
Option: 3 $17$
Option: 4 $16$

What is Ellipse? -

Ellipse

$\mathbf{\frac{\mathbf{x}^{2}}{\mathbf{a}^{2}}+\frac{\mathbf{y}^{2}}{\mathbf{b}^{2}}=1} \quad \text { where }, \mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)_{(\mathrm{a}>\mathrm{b})}$

a > b
the length of the major axis is 2a
the coordinates of the vertices are (±a, 0)
the length of the minor axis is 2b
the coordinates of the co-vertices are (0, ±b)

-

What is Hyperbola? -

Hyperbola:

Eccentricity of Hyperbola:

$\\\mathrm{Equation\;of\;the\;hyperbola\;is\;\;\;\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}\\\text{we have,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;b^2=a^2\left ( e^2-1 \right )}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e^2=\frac{b^2+a^2}{a^2}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e=\sqrt{1+\left ( \frac{b^2}{a^2} \right )}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e=\sqrt{1+\left ( \frac{2b}{2a} \right )^2}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e=\sqrt{1+\left ( \frac{conjugate \;axis}{transverse \;axis} \right )^2}}$

-

$\begin{aligned} &e_{1}=\sqrt{1-\frac{4}{18}}=\sqrt{\frac{7}{9}}=\frac{\sqrt{7}}{3}\\ &\mathrm{e}_{2}=\sqrt{1+\frac{4}{9}}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}\\ &15 e_{1}^{2}+3 e_{2}^{2}=k \Rightarrow \quad k=15\left(\frac{7}{9}\right)+3\left(\frac{13}{9}\right) \end{aligned}$

So, k= 16

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Posted by

avinash.dongre

Let C the centroid of the triangle with vertices $(3,-1),(1,3)\: and \: (2,4).$ Let P be the point of intersection of the lines $x+3y-1 =0$ and $3x-y+1 =0.$ Then the line passing through the points C and P also passes through the point:
Option: 1 $(-9,-7)$
Option: 2 $(-9,-6)$
Option: 3 $(7,6)$
Option: 4 $(9,7)$

Centroid -

Centroid

Centroid of a triangle is the point of intersection of the medians of the triangle. A centroid divides the median in the ratio 2:1.

Whereas, the median is the line joining the mid-points of the sides and the opposite vertices.

The coordinates of the centroid of a triangle (G) whose vertices are A (x₁, y₁), B (x₂, y₂) and C(x₃, y₃), is given by

$\\\mathrm{\mathbf{\left ( \frac{x_1+x_2+x_3}{3},\;\frac{y_1+y_2+y_3}{3} \right )}}$

If D (a₁, b₁), E (a₂, b₂) and F (a₃, b₃) are the mid point of ΔABC, then its centroid is given by

$\\\mathrm{\mathbf{\left ( \frac{a_1+a_2+a_3}{3},\;\frac{b_1+b_2+b_3}{3} \right )}}$

-

Point of intersection of two lines -

Point of intersection of two lines

Equation of two non-parallel line is

$\\L_1=a_1x+b_1y+c_1=0\\L_2=a_2x+b_2y+c_2=0$

If P (x₁, y₁) is a point of intersection of L₁ and L₂ , then solving these two equations of the line by cross multiplication

$\frac{x_1}{b_1c_2-c_1b_2}=\frac{y_1}{c_1a_2-a_1c_2}=\frac{1}{a_1b_2-b_1a_2}$

We get,

$\mathbf{\left ( x_1,y_1 \right )=\left ( \frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1} \right )}$

-

Equation of Straight Line (Part 2) -

Equation of Straight Line

(c) Two-point form

The equation of a straight line passing through the two given points (x₁,y₁) and (x₁,y₁)is given by

.

-

The centroid of triangle ABC D(2,2)

Point of intersection P $\left ( -\frac{1}{5},\frac{2}{5} \right )$

equation of line DP is 8x – 11y + 6 = 0

Point (–9,–6) satisfies the equation

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Posted by

avinash.dongre

Among the following compounds, the increasing order of their basic strength is
Option: 1 (I) < (II) < (IV) < (III)
Option: 2 (I) < (II) < (III) < (IV)
Option: 3 (II) < (I) < (IV) < (III)
Option: 4 (II) < (I) < (III) < (IV)

Among the following compounds, the increasing order of their basic strength is (II)<(I)<(IV)<(III)
(III) is most basic due to +I effect of −CH₃ group attached to N. This increases electron density on N so that N can easily donate its lone pair of electrons to suitable acid.
(II) is least basic as the lone pair of electrons on N is part of pi electrons of aromatic ring. Donation of this lone pair of electrons will break the stability of the aromatic ring.
(I) is less basic than (IV) as the lone pair of electrons on N is in conjugation with aromatic ring.

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Posted by

vishal kumar

The increasing order of the boiling points for the following compounds is :
Option: 1 (III) < (IV) < (II) < (I)
Option: 2 (IV) < (III) < (I) < (II)
Option: 3 (II) < (III) < (IV) < (I)
Option: 4 (III) < (II) < (I) < (IV)

The increasing order of boiling point is

$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OCH}_{3}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}_{3}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$

in C2?H5?OH (i) C-O bond is polar & form hydrogen bonding in H2?O

in C2?H5?Cl → C-Cl bond is polar but less than that of C-O bond
so boiling point of C2?H5?Cl is less than that of C2?H5?OH
in C2?H5?OCH3? , the C-O bond is polar but the weak polarity of ethers do not appreciably affect their boiling points

And alkane has the least boiling point than above compounds.

Answer: (1)

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Posted by

vishal kumar

A compound of molecular formula C₈H₈O₂ reacts with acetophenone to form a single cross-aldol product in the presence of base. The same compound on reaction with conc. NaOH forms benzyl alcohol as one of the products. The structure of the compound is :
Option: 1

Option: 2

Option: 3

Option: 4

The structure of the compound is :

Correct option 1

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Posted by

vishal kumar

The tangent at the point (2, −2) to the curve, x²y²−2x=4(1−y) does not pass through the point :
Option: 1 (8,5)
Option: 2 (4,1/3)
Option: 3 (-2,-7)
Option: 4 (-4,-9)

As learnt in concept ABCD

$x^{2}y^{2}-2x=4(1-y)$

Differentiate both sides wrt.x.

$2xy^{y}+x^{2}.2yy{}'-2=4-4y{}'$

$y{}'=-\frac{(xy^{2}-1)}{2+x^{2}y}=\left ( \frac{1-xy^{2}}{2+x^{2}y} \right )$

at (2,-2) $y{}'=\frac{1-2\times (-2)^{2}}{2+2^{2}\times (-2)}=\frac{1-8}{2-8}=+\frac{7}{6}$

Equation of tangent is

$\frac{y+2}{x-2}=\frac{7}{6}=>7x-6y-26=0$

It doesn't pass through (-2,-7)

Concept ABCD

Slope of curve at a given point

To find slope, we differentiate with respect to x and find $\frac{dy}{dx}$

Eg in $x^{2}-y^{2}=4$

$\frac{dy}{dx}=\frac{x}{y}$

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vishal kumar

A flask contains a mixture of isohexane and 3-methylpentane. One of the liquids boils at $\mathrm{63 \degree C}$ while the other boils at $\mathrm{60 \degree C}$ . What is the best way to separate the two liquids and which one will be distilled out first?
Option: 1 Fractional distillation, isohexane
Option: 2 simple distillation,3-methylpentane
Option: 3 fractional distillation, 3 - methylpentane
Option: 4 simple distillation, isohexane

As we have learnt,

Distillation and fractional distillation -

Simple distillation is a procedure by which two liquids with different boiling points can be separated. Simple distillation can be used effectively to separate liquids that have some major degrees difference in their boiling points. As the liquid being distilled is heated, the vapours that form will be richest in the component of the mixture that boils at the lowest temperature. Purified compounds will boil, and thus turn into vapours, over a relatively small temperature range (2 or 3°C) by carefully watching the temperature in the distillation flask, it is possible to affect a reasonably good separation.

As distillation progresses, the concentration of the lowest boiling component will steadily decrease. Eventually, the temperature within the apparatus will begin to change; a pure compound is no longer being distilled.

Steps to be followed in simple distillation process:

(i) Take a mixture(Acetone and water) in the distillation flask fit it with the thermometer.
(ii) Arrange the apparatus as shown in the given figure.
(iii) Heat the mixture slowly keeping a close watch on thermometer.
(iv) Since the acetone has lower boiling point starts vaporises and condenses in the condenser which is finally collected in the beaker.

Note: Simple distillation is effective only when separating a volatile liquid from a nonvolatile substance. If the liquids comprising the mixture that is being distilled have boiling points that are closer than 50 degrees to one another, the distillate collected will be richer in the more volatile compound but not to the degree necessary for complete separation of the individual compounds.

Fractional distillation is used for the mixture of two liquids which differ in their boiling point by 10-15 K. It is a type of distillation which involves the separation of miscible liquids. The process involves repeated distillations and condensations and the mixture is usually separated into component parts. In this method a fractionating column is used to increase the cooling surface area so that ascending vapour phase become richer in more volatile component and descending liquid phasebecome richer in less volatile component.

The basic principle of this type of distillation is that different liquids boil and evaporate at different temperatures. So when the mixture is heated, the substance with lower boiling point starts to boil first and convert into vapours.

Fractinal distillation is used for seperation of

(i) Acetone (b.pt. 329 K) from methyl alcohol (b.pt. 338 K)

(ii) Crude oil into various useful fractions such as gasoline, kerosene oil, lubricating oil, etc.

Steps to be followed in fractional distillation process:

1. After setting up the apparatus, a mixture of two miscible liquids A and B is taken where A has more volatility than substance B.

2. The solution is added into the distilling flask while the fractionating column is connected at the tip of the flask. Heat is applied which increases the temperature slowly. The mixture then starts to boil and vapours start rising in the flask.

3. The vapours are from the volatile component A. The vapours then start moving through the fractionating column into the condenser where it is cooled down to form a liquid which is collected in the receiver.

Throughout the process, vaporization and condensation take place repeatedly until the two mixtures are separated completely.

-

Liquid having lower boiling point comes out first in fractional distillation. Simple distillation can't be used as boiling point difference is very small.

Therefore, Option(1) is correct.

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Posted by

vishal kumar

Let the normal at a point P on the curve $y^{2}-3x^{2}+y+10=0$ intersect the y-axis at $\left ( 0,\frac{3}{2} \right ).$ If m is the slope of the tangent at P to the curve, then $\left | m \right |$ is equal to
Option: 1 4
Option: 2 3
Option: 3 2
Option: 4 1

Equation of Straight Line (Part 1) -

Equation of Straight Line

(b) Point-Slope form

Let the equation of give line l with slope ‘m’ is

y = mx + c …..(i)

(x₁,y₁) lies on the line i

y₁= mx₁+c ……(ii)

From (i) and (ii) [(ii) - (i)]

y - y₁= m( x - x₁)

The equation of a straight line whose slope is given as ‘m’ and passes through the point (x₁,y₁) is .

-

$\begin{array}{c}{2 y y^{\prime}+y^{\prime}-6 x=0} \\ {y^{\prime}=\frac{6 x}{2 y+1}} \\ {\frac{-1}{y^{\prime}}=\frac{-(2 y+1)}{6 x}}(\text{slope of the normal})\end{array}$

Equation of the normal $y-y_{1}=\frac{-\left(2 y_{1}+1\right)}{6 x_{1}}\left(x-x_{1}\right)$

Normal intersect at (0,3/2)

$\\\frac{3}{2}-y_{1}=\frac{-\left(2 y_{1}+1\right)}{6 x_{1}}\left(0-x_{1}\right)\\ 8y_1-8=0\\ y_1=1\\ x_1=\pm2\\ |m|=4$

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All Questions

If one end of a focal chord AB of the parabola is at then the equation of the tangent to it at B is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Shabareesh

The length of the minor axis (along y-axis) of an ellipse in the standard form is If this ellipse touches the line, then its eccentricity is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

Crack CUET with india's "Best Teachers"

If are the eccentricities of the ellipse, and the hyperbola, respectively and is a point on the ellipse, then is equal to : Option: 1 Option: 2 Option: 3 Option: 4

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avinash.dongre

Let C the centroid of the triangle with vertices Let P be the point of intersection of the lines and Then the line passing through the points C and P also passes through the point: Option: 1 Option: 2 Option: 3 Option: 4

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avinash.dongre

JEE Main high-scoring chapters and topics

Among the following compounds, the increasing order of their basic strength is Option: 1 (I) < (II) < (IV) < (III) Option: 2 (I) < (II) < (III) < (IV) Option: 3 (II) < (I) < (IV) < (III) Option: 4 (II) < (I) < (III) < (IV)

If one end of a focal chord AB of the parabola $y^{2}=8x$ is at $A\left ( \frac{1}{2},-2 \right ),$ then the equation of the tangent to it at B is :
Option: 1 $x+2y+8=0$
Option: 2 $2x-y-24=0$
Option: 3 $x-2y+8=0$
Option: 4 $2x+y-24=0$

Among the following compounds, the increasing order of their basic strength is
Option: 1 (I) < (II) < (IV) < (III)
Option: 2 (I) < (II) < (III) < (IV)
Option: 3 (II) < (I) < (IV) < (III)
Option: 4 (II) < (I) < (III) < (IV)

The increasing order of the boiling points for the following compounds is :
Option: 1 (III) < (IV) < (II) < (I)
Option: 2 (IV) < (III) < (I) < (II)
Option: 3 (II) < (III) < (IV) < (I)
Option: 4 (III) < (II) < (I) < (IV)

The tangent at the point (2, −2) to the curve, x²y²−2x=4(1−y) does not pass through the point :
Option: 1 (8,5)
Option: 2 (4,1/3)
Option: 3 (-2,-7)
Option: 4 (-4,-9)

Let the normal at a point P on the curve $y^{2}-3x^{2}+y+10=0$ intersect the y-axis at $\left ( 0,\frac{3}{2} \right ).$ If m is the slope of the tangent at P to the curve, then $\left | m \right |$ is equal to
Option: 1 4
Option: 2 3
Option: 3 2
Option: 4 1