JEE Main 2021 — Mathematics Coordinate Geometry

Given curves, $\frac{{x}^{2}}{a}+\frac{{y}^{2}}{b}=1.....(i)$ax2​+by2​=1.....(i)

On differentiating equation $(i)$(i), we get

$$\frac{2x}{a}+\frac{2y}{b}\frac{dy}{dx}=0$$a2x​+b2y​dxdy​=0

$$\Rightarrow {m}_{1}=\frac{dy}{dx}=-\frac{xb}{ya}...(iii)$$⇒m1​=dxdy​=−yaxb​...(iii)

and $\frac{{x}^{2}}{c}+\frac{{y}^{2}}{d}=1.....(ii)$cx2​+dy2​=1.....(ii)

On differentiating equation $(ii)$(ii), we get

$$\frac{2x}{c}+\frac{2y}{d}\frac{dy}{dx}=0$$c2x​+d2y​dxdy​=0

$$\Rightarrow {m}_{2}=\frac{dy}{dx}=-\frac{xd}{yc}...(iv)$$⇒m2​=dxdy​=−ycxd​...(iv)

For orthogonal curves, ${m}_{1}\times {m}_{2}=-1$m1​×m2​=−1

$$\Rightarrow (-\frac{xb}{ya})\times (-\frac{xd}{yc})=-1$$⇒(−yaxb​)×(−ycxd​)=−1

$$\frac{{x}^{2}}{{y}^{2}}=-\frac{ac}{bd}....(v)$$y2x2​=−bdac​....(v)

From equation $(i)-(ii)$(i)−(ii), we get

$${x}^{2}(\frac{c-a}{ac})={y}^{2}(\frac{b-d}{bd})$$x2(acc−a​)=y2(bdb−d​)

$$\Rightarrow \frac{{x}^{2}}{{y}^{2}}=(\frac{b-d}{c-a})(\frac{ac}{bd})$$⇒y2x2​=(c−ab−d​)(bdac​)

From equation $(v)$(v), we get

$$a-c=b-d$$a−c=b−d

$\Rightarrow a-b=c-d$⇒a−b=c−d.