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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 $2x+y-24=0$

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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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avinash.dongre

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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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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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avinash.dongre

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If the image of the point P(1, −2, 3) in the plane, 2x+3y−4z+22=0 measured parallel to the line, $\frac{x}{1}= \frac{y}{4}= \frac{z}{5}$ is Q, then PQ is equal to :
Option: 1 $2\sqrt{42}$

Option: 5 $\sqrt{42}$

Option: 9 $6\sqrt{5}$

Option: 13 $3\sqrt{5}$

Intersection of line and plane -

Let the line

$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$ plane

$a_{1}x+b_{1}y+c_{1}z+d=0$ intersect at P

to find P assume general point on line as $\left ( x_{1}+\lambda a_{1}y_{1} +\lambda b_{1}z_{1}+\lambda c_{1}\right )$

now put it in plane to find $\lambda$ ,

$a_{1}\left ( x_{1}+\lambda a \right )+b_{1}\left ( y_{1}+\lambda b \right )+c_{1}\left ( z_{1}+\lambda c \right )+d=0$

Distance formula -

The distance between two points $A(x_{1},y_{1},z_{1})\, and \, B(x_{2},y_{2},z_{2})$ is = $\sqrt{\left ( x_{2}-x_{1} \right )^{2}+{\left ( y_{2}-y_{1} \right )^{2}}+{\left ( z_{2}-z_{1} \right )^{2}}}$

- wherein

$\vec{OA}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}$

$\vec{OB}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}$

$\underset{AB}{\rightarrow}$ = $\underset{OB}{\rightarrow}$ - $\underset{AB}{\rightarrow}$

$\underset{AB}{\rightarrow}$ $= \left ( x_{2}-x_{1} \right )\hat{i}+\left ( y_{2}-y_{1} \right )\hat{j}+\left ( z_{2}-z_{1} \right )\hat{k}$

$AB= \left | \underset{AB}{\rightarrow} \right |$

$AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}$

PQ=2PM

Now, for finding M

$\frac{x}{1}=\frac{y}{4}=\frac{z}{5}=k(say)$

$\Rightarrow x=k,y=4k,z=5k$

Putting in equation of plane,

$k=\frac{11}{3}$

So, $M=\left (\frac{11}{3} ,\frac{44}{3},\frac{55}{3} \right )$

$\therefore PM=\sqrt{\frac{64}{9}+\frac{2500}{9}+\frac{2116}{9}}= 2\sqrt{130}$

$\Rightarrow PQ=4\sqrt{130}$

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Posted by

vishal kumar

The line of intersection of the planes $\small \underset{r}{\rightarrow} .\left ( 3\hat{i} -\hat{j} +\hat{k}\right )= 1$ and $\small \underset{r}{\rightarrow} .\left ( \hat{i} +4 \hat{j} -2\hat{k}\right ) =2$ , is:
Option: 1 $-2 \hat{i}+7 \hat{j}+13 \hat{k}$
Option: 2 $2 \hat{i}+7 \hat{j}-13 \hat{k}$
Option: 3 $-2 \hat{i}-7 \hat{j}+13 \hat{k}$
Option: 4 $-2 \hat{i}+7 \hat{j}+13 \hat{k}$

As we have learned

Equation of line as intersection of two planes -

Let the two intersecting planes be

$ax+by+cz+d= 0$ and

$a_{1}x+b_{1}y+c_{1}z+d_{1}= 0$

then the parallel vector of line formed their intersection can be obtained by

$\begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ a&b &c \\ a_{1} & b_{1} & c_{1} \end{vmatrix}= A\hat{i}+B\hat{j}+C\hat{k}(assumed)$

and points can be obtained by putting $z= 0$ and solving

$ax+by+d= 0$ and

$a_{1}x+b_{1}y+d_{1}= 0$ say $\alpha ,\beta$

Now the equation will be

$\frac{x-\alpha }{A}=\frac{y-\beta }{B}=\frac{z-0 }{C}$

$3x-y + z = 1 \\ and \: \: x + 4y - 2z = 2$

putting z = 0

3x-y = 1 and x + 4y = 2

$\Rightarrow 13 x = 6 \Rightarrow x = 6/13 \\ y = 5/3$

$vector \: \: along \: \: line = \begin{vmatrix} \hat i & \hat j &\hat k \\ 3 & -1 & 1\\ 1 & 4 &-2 \end{vmatrix} = -2 \hat i + 7 \hat j +13 \hat k$

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Posted by

vishal kumar

NEET 2024 Most scoring concepts

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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

The strength of an aqueous NaOH solution is most accurately determined by titrating:
(Note: consider that an appropriate indicator is used)
Option: 1 Aq.NaOH in a pipette and aqueous oxalic acid in a burette
Option: 2 Aq.NaOH in a volumetric flask and concentrated $H_{2}SO_{4}$ in a conical flask
Option: 3 Aq.NaOH in a burette and concentrated $H_{2}SO_{4}$ in a conical flask
Option: 4 Aq.NaOH in a burette and aqueous oxalic acid in a conical flask

As we have learnt,

To calculate the strength of NaOH, it is titrated against oxalic acid. For this purpose, NaOH is kept in a burette and oxalic acid in a conical flask.

Therefore, Option(4) is correct.

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vishal kumar

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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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Kuldeep Maurya

Let the line $y=mx$ and the ellipse $2x^{2}+y^{2}=1$ intersect at a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at $\left ( -\frac{1}{3\sqrt{2}},0 \right )$ and $(0,\beta )$ , then $\beta$ is equal to :
Option: 1 $\frac{2}{\sqrt{3}}$
Option: 2 $\frac{2}{3}$
Option: 3 $\frac{2\sqrt{2}}{3}$
Option: 4 $\frac{\sqrt{2}}{3}$

Equation of Normal in Point Form and Parametric Form -

Equation of Normal in Point Form and Parametric Form

Point form:
$\\ {\text {The equation of normal at }\left(x_{1}, y_{1}\right) \text { to the ellipse, } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { is }} \\ {\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}=a^{2}-b^{2}.}$

$\\ {\text { Let } P \text { be }\left(x_{1}, y_{1}\right)} \\\\ {\text { Equation of normal at } P \text { is } \frac{x}{2 x_{1}}-\frac{y}{y_{1}}=-\frac{1}{2}} \\\\ {\text { It passes through }\left(-\frac{1}{3 \sqrt{2}}, 0\right) \Rightarrow \frac{-1}{6 \sqrt{2} x_{1}}=-\frac{1}{2} \Rightarrow x_{1}=\frac{1}{3 \sqrt{2}}}$

$\\ {\text { So } y_{1}=\frac{2 \sqrt{2}}{3} \text { (as } P \text { lies in } 1 \text { 'quadrant) }} \\\\ {\text { So } \beta=\frac{y_{1}}{2}=\frac{\sqrt{2}}{3}}$

Correct option (4)

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Kuldeep Maurya

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Get Answers to all your Questions

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

Posted by

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

Posted by

avinash.dongre

JEE Main high-scoring chapters and topics

If the image of the point P(1, −2, 3) in the plane, 2x+3y−4z+22=0 measured parallel to the line, is Q, then PQ is equal to : Option: 1 Option: 5 Option: 9 Option: 13

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 $2x+y-24=0$

If the image of the point P(1, −2, 3) in the plane, 2x+3y−4z+22=0 measured parallel to the line, $\frac{x}{1}= \frac{y}{4}= \frac{z}{5}$ is Q, then PQ is equal to :
Option: 1 $2\sqrt{42}$

Option: 5 $\sqrt{42}$

Option: 9 $6\sqrt{5}$

Option: 13 $3\sqrt{5}$

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