Question

Let the normals at the four points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ and $(x_4,y_4)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be concurrent at some point (called as conormal point). Then $(x_1+x_2+x_3+x_4)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4})$ is equal to

• A. $4$
• B. $3$
• C. $-4$
• D. $2$

Answer 1

The correct option is A $4$

Let the point of concurrency be $(h,k)$
The equation of normal at $(x^{\prime },y^{\prime })$ is $\frac{x-x^{\prime }}{\frac{x^{\prime }}{a^2}}=\frac{y-y^{\prime }}{\frac{y^{\prime }}{b^2}}$
It passes through $(h,k)$
Thus, $(a^2-b^2)x^{\prime }{y}^{\prime }+b^{2}k{x}^{\prime }-a^{2}h{y}^{\prime }=0\cdots \left(1\right)$
Equation $\left(1\right)$ is satisfied by $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ and $(x_4,y_4)$
Thus the curve passing through these points is $(a^2-b^2)xy+b^{2}kx-a^{2}hy=0$
From equation $\left(1\right)$,
$y^{\prime }=\frac{b^{2}k{x}^{\prime }}{a^{2}h-(a^2-b^2)x^{\prime }}\cdots \left(2\right)$
But, $\frac{x^{\prime 2}}{a^2}+\frac{y^{\prime 2}}{b^2}=1\cdots \left(3\right)$
Substituting equation $\left(2\right)$ in equation $\left(3\right),$ Thus eliminating $y^{\prime }$ we get,
$-(a^2-b^2)x^{\prime 4}+2h{a}^2(a^2-b^2)x^{\prime 3}+\cdots -2a^{4}h(a^2-b^2)x^{\prime }+a^6{h}^2=0$
Then $x_1+x_2+x_3+x_4=\frac{2h{a}^2}{a^2-b^2}$
$x_1{x}_2{x}_3+x_2{x}_3{x}_4+x_3{x}_4{x}_1+x_1{x}_2{x}_4=-2a^{4}h$
and, $x_1{x}_2{x}_3{x}_4=\frac{a^6{h}^2}{-(a^2-b^2)}$
So,$(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4})=\frac{2(a^2-b^2)}{a^{2}h}$
Hence, $(x_1+x_2+x_3+x_4)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4})=4$