Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Applications of Derivatives

Question:

The curves $ax^2+by^2=1$ and $Ax^2+By^2=1$ intersect orthogonally, then:

Options:

$\frac{1}{a}+\frac{1}{A}=\frac{1}{b}+\frac{1}{B}$

$\frac{1}{a}-\frac{1}{A}=\frac{1}{b}-\frac{1}{B}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{B}-\frac{1}{A}$

None of these

Correct Answer:

$\frac{1}{a}-\frac{1}{A}=\frac{1}{b}-\frac{1}{B}$

Explanation:

$ax^2+by^2=Ax^2+By^2⇒(a-A)x^2=(B_b)y^2⇒\frac{x^2}{y^2}=\frac{B-b}{a-A}$

$ax^2+by^2=1⇒\frac{dy}{dx}=\frac{-ax}{by}=m_1$

$Ax^2+By^2=1⇒\frac{dy}{dx}=\frac{-Ax}{By}=m_2$

$m_1.m_2=-1⇒\frac{a.A}{b.B}.\frac{x^2}{y^2}=-1⇒\frac{a.A}{b.B}.(\frac{B-b}{a_A})=-1$

$⇒\frac{B-b}{b.B}=\frac{A-a}{aA}⇒\frac{1}{a}-\frac{1}{A}=\frac{1}{b}-\frac{1}{B}$