Question

Product of the perpendicular from the foci upon any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a<b)$ is equal to.

• A. $2a$
• B. $a^2$
• C. $b^2$
• D. $a{b}^2$

Answer 1

Answer: (C)

$b^2$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Let $P$ point $\left(x_0{y}_0\right)$

$\frac{x_{0}x}{a^2}+\frac{y_{0}y}{b^2}=1$ tangent focus $(c,0)(-c,0)$

$d=\frac{\frac{x_{0}c}{a^2}\pm 1}{\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}}}\Rightarrow K=\frac{\frac{x^2{c}^2}{a^4}-1}{\frac{x_0^2}{a^4}+\frac{y_0^2}{b^4}}$

$\Rightarrow \frac{\frac{x_0^2{c}^2-a^4}{a^4}}{\frac{x_0^2{a}^4+y_0^2{a}^4}{a^4{b}^4}}=\frac{(x^2{c}^2-a^4)\left(a^4{b}^4\right)}{a^4(x_0^2{b}^4+y_0^2{a}^4)}$

we know that

$\Rightarrow \frac{(x^2{c}^2-a^4)b^4}{x_0^2{b}^4+y_0^2{a}^4}\Rightarrow y_0^2{a}^4=a^4{b}^2-a^2{b}^2{x}_0^2$

$\Rightarrow \frac{(x_0^2{c}^2-a^4)b^4}{x_0^2{b}^4+a^4{b}^2-a^2{b}^2{x}_0^2}\Rightarrow \frac{b^2{x}^2{c}^2-a^4}{x_0^2(b^2-a^2)+a^4}$

$\Rightarrow \frac{b^2(x_0^2{c}^2-a^4)}{x_0^2(-c^2)+a^4}=\frac{b^2(x_0^2{c}^2-a^4)}{-(x_0^2{c}^2-a^4)}$

$k=-b^2$

distance $=b^2$