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S₁ : If from a point $P(0,\alpha )$ two normal other than axes are drawn to ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1,$ where $\left | \alpha \right |\leq k,$ then least value of k is $\frac{9}{4}$

S₂ : The minimum and maximum distances of a point $(1, 2)$ from the ellipse $4x^{2} + 9y^{2} + 8x - 36y + 4 = 0$ are L and G, then G – L is equal to 4

S₃ : If the length of latus rectum of an ellipse is one-third of its major axis. Its eccentricity is equal to $\frac{2}{3}$

S_{4 :}The set of all positive values of a for which $(13x-1)^{2}+(13y-2)^{2}=\left ( \frac{5x+12y-1}{a} \right )^{2}$ represents an ellipse is $(1, 2)$

Option: 1
$TTFF$

Option: 2
$TTFT$

Option: 3
$TTTF$

Option: 4
$TFTF$

Answers (1)

best_answer

Standard equation -

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$

- wherein

$a\rightarrow$ Semi major axis

$b\rightarrow$ Semi minor axis

Eccentricity -

$e= \sqrt{1-\frac{b^{2}}{a^{2}}}$

- wherein

For the ellipse

$\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1$

S₁ : Equation normal at $P(\Theta )$ is $5\; \; \; \; \sec \: \Theta \: x-4\: \cos \sec \Theta \: \; \; \; \; \; \; \; y=25-16$ and it passes through $P(0,\alpha )$

$\alpha =\frac{-9}{4\cos \sec \Theta }$ i.e. $\alpha =\frac{-9}{4}\sin \Theta \:\:\:\:\:\:\left | \alpha \right |\leq \frac{9}{4}$

S₂ : $4x^{2}+8x+9y^{2}-36y=-4$

$4(x^{2}+2x+1)+9(y^{2}-4y+4)=36$

$\frac{(x+1)^{2}}{9}+\frac{(y-2)^{2}}{4}=1$

$\therefore G=5,L=1$

S₃ : $\frac{2b^{2}}{a}=\frac{2a}{3}\Rightarrow 3b^{2}=a^{2}$

$\therefore b^{2}=a^{2}(1-e^{2})\Rightarrow 1=3(1-e^{2})\Rightarrow e=\sqrt{2/3}$

S₄ : The equation is $\left ( x-\frac{1}{13} \right )^{2}+\left ( y-\frac{2}{13} \right )^{2}$

$=\frac{1}{a^{2}}\left ( \frac{5x+12y-1}{13} \right )^{2}$

$\therefore \frac{1}{a^{2}}<1$ i.e. $a^{2}>1$

Posted by

Sanket Gandhi

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