Tangents are drawn from point P(3,4) to ellipse frac{x^2}{9}+frac{y^2}{4}=1 touching ellipse at poin

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10-2. Parabola, Ellipse, Hyperbola

normal

Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $\mathrm{A}$ and $\mathrm{B}$.

$1.$ The coordinates of $\mathrm{A}$ and $\mathrm{B}$ are

$(A)$ $(3,0)$ and $(0,2)$

$(B)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

$(C)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $(0,2)$

$(D)$ $(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

$2.$ The orthocentre of the triangle $\mathrm{PAB}$ is

$(A)$ $\left(5, \frac{8}{7}\right)$ $(B)$ $\left(\frac{7}{5}, \frac{25}{8}\right)$

$(C)$ $\left(\frac{11}{5}, \frac{8}{5}\right)$ $(D)$ $\left(\frac{8}{25}, \frac{7}{5}\right)$

$3.$ The equation of the locus of the point whose distances from the point $\mathrm{P}$ and the line $\mathrm{AB}$ are equal, is

$(A)$ $9 x^2+y^2-6 x y-54 x-62 y+241=0$

$(B)$ $x^2+9 y^2+6 x y-54 x+62 y-241=0$

$(C)$ $9 x^2+9 y^2-6 x y-54 x-62 y-241=0$

$(D)$ $x^2+y^2-2 x y+27 x+31 y-120=0$

Give the answer question $1,2$ and $3.$

$(A,B,C)$

$(D,C,A)$

$(B,B,D)$

$(A,A,C)$

(IIT-2010)

Solution

$1.$ $ y=m x+\sqrt{9 m^2+4} $

$ 4-3 m=\sqrt{9 m^2+4} $

$ 16+9 m^2-24 m=9 m^2+4 \Rightarrow m=\frac{12}{24}=\frac{1}{2}$

Equation is $y-4=\frac{1}{2}(x-3)$

$2 \mathrm{y}-8=\mathrm{x}-3 \Rightarrow \mathrm{x}-2 \mathrm{y}+5=0$

Let $\mathrm{B}=(\alpha, \beta) \Rightarrow \frac{x \alpha}{9}+\frac{y \beta}{4}-1=0 \Rightarrow \frac{\alpha / 9}{1}=\frac{\beta / 4}{-2}=\frac{-1}{5} \Rightarrow \alpha=-\frac{9}{5}, \beta=\frac{8}{5}$

$B \equiv\left(-\frac{9}{5}, \frac{8}{5}\right)$

$2.$ Slope of $BD$ must be $0$

$\Rightarrow y-\frac{8}{5}=0 \quad\left(x+\frac{9}{5}\right) \Rightarrow y=\frac{8}{5}$

Hence $y$ coordinate of $D$ is $8 / 5$.

$3.$ Locus is parabola

Equation of $\mathrm{AB}$ Is $\frac{3 x}{9}+\frac{4 y}{4}=1 \Rightarrow \frac{x}{3}+y=1 \Rightarrow \mathrm{x}+3 \mathrm{y}-3=0$

$ (x-3)^2+(y-4)^2=\frac{(x+3 y-3)^2}{10} $

$ 10 x^2+90-60 x+10 y^2+160-80 y=x^2+9 y^2+9+6 x y-6 x-18 y $

$ \Rightarrow 9 x^2+y^2-6 x y-54 x-62 y+241=0$

Standard 11

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