Question

Find the condition that the curves $\frac{x^2}{a}+\frac{y^2}{b}=1$ & $\frac{x^2}{a^{\prime }}+\frac{y^2}{b^{\prime }}=1$ may cut orthogonally

• A. $a-b^{\prime }=a^{\prime }-b$
• B. $a^{\prime }-b=a-b^{\prime }$
• C. $a-b=a^{\prime }-b^{\prime }$
• D. $-a+b=a^{\prime }-b^{\prime }$

Answer 1

The correct option is C $a-b=a^{\prime }-b^{\prime }$

$\frac{x^2}{a}+\frac{y^2}{b}=1\Rightarrow m_1=\frac{-bx}{ay}$ ...(1)
$\frac{x^2}{a_1}+\frac{y^2}{b_1}=1\Rightarrow m_2=\frac{-b_{1}x}{a_{1}y}$ ...(2)
Solving them we get $x^2=a_1=y^2=b$
Now $m_1{m}_2\frac{b{b}_1{a}_1}{a{a}_1{b}_1}=\frac{b_1}{a_1}=-1\therefore a_1{+b}_2=a+b$