Question

If the line $x\cos a+y\sin a=p$ be normal to the ellipse $\frac{x^2}{a^2}$ $+\frac{y^2}{b^2}$ = 1 then

• A. $p^2(a^2{\cos }^{2}a+b^2{\sin }^{2}a)=a^2-b^2$
• B. $p^2(a^2{\cos }^{2}a+b^2{\sin }^{2}a)=(a^2-b^2{)}^2$
• C. $p^2(a^2{\sec }^{2}a+b^2{\csc }^{2}a)=(a^2-b^2)$
• D. $p^2(a^2{\sec }^{2}a+b^2{\csc }^{2}a)=(a^2-b^2{)}^2$

Answer 1

Answer: (A)

$p^2(a^2{\cos }^{2}a+b^2{\sin }^{2}a)=a^2-b^2$

A line $y=mx+c$ is normal to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ if $c^2=m^2\frac{(a^2-b^2{)}^2}{a^2+m^2{b}^2}$

Given equation $x\cos a+y\sin a=p\Rightarrow y=-\left(\frac{\cos a}{\sin a}\right)x+\frac{p}{\sin a}$

Here $m=-\cot a,c=\frac{p}{\sin a}$

Substituting in the formulae, we get

$\frac{p^2}{{\sin }^{2}a}=\frac{{\cos }^{2}a}{si{n}^{2}a}\times \frac{(a^2-b^2{)}^2}{a^2+\left({\cot }^{2}a\right)b^2}$

After simplification, we get

$p^2\frac{(a^2{\sin }^{2}a+b^2{\cos }^{2}a)}{{\sin }^{2}a{\cos }^{2}a}=(a^2-b^2{)}^2$

$p^2(a^2{\sec }^{2}a+b^2{\csc }^{2}a)=(a^2-b^2{)}^2$