Question

If 4x² + py² = 45 and x² - 4y² = 5 cut orthogonally, then the value of p is:

• (A) $9$
• (B) $\frac{1}{3}$
• (C) $3$
• (D) $18$

Answer 1

The Correct Option is A

Explanation:
Given:Equations of curves,  4x² + py² = 45 and x² - 4y² = 5 cut orthogonally.We have to find the value of p.Consider,4x² + py² = 45Differentiating w.r.t x, we get$8x+2$ py $\frac{dy}{dx}=0[\text{ Using standard derivatives- }1]$$\Rightarrow \frac{dy}{dx}=\frac{-4x}{py}$Similarly for $x^2-4y^2=5$$\frac{dy}{dx}=\frac{x}{4y}$Let $(\alpha ,\beta )$ be the point of contact.Then, $4{\alpha }^2+p{\beta }^2=45$...(i)${\alpha }^2-4{\beta }^2=5$ ....(ii)Multiplying (ii) by 4 and subtracting it from (i), we get$(p+16){\beta }^2=25$$\Rightarrow {\beta }^2=\frac{25}{p+16}$...(iii)Using (ii), we get${\alpha }^2=5+4\cdot \frac{25}{p+16}$
$=\frac{5p+80+100}{p+16}$
$=\frac{5p+180}{p+16}$
$=\frac{5(p+36)}{p+16}$Dividing (iv) by (iii), we get$\frac{{\alpha }^2}{{\beta }^2}=\frac{p+36}{5}$Now, using tangents and normals, we get$m_1={\left(\frac{dy}{dx}\right)}_{(\alpha ,\beta )}=\frac{4\alpha }{p\beta }$and $m_2={\left(\frac{dy}{dx}\right)}_{(\alpha ,\beta )}=\frac{\alpha }{4\beta }$: Both curves cut orthogonally, then${\mathrm{m}}_1{\mathrm{m}}_2=-1$$\Rightarrow \left(\frac{-4\alpha }{p\beta }\right)\cdot \frac{\alpha }{4\beta }=-1$$\Rightarrow \frac{1}{p}\left(\frac{p+36}{5}\right)=1$$\Rightarrow 5p=p+36$$\Rightarrow p=9$Hence, the correct option is (A).