Question

Find the point on the curve 3x² – 4y² = 72 which is nearest to the line 3x + 2y + 1 = 0

• A. (–6, –3)
• B. (–6, 3)
• C. (3, –6)
• D. None of these

Answer 1

Answer: (B)

(–6, 3)

Answer 2

Slope of the given line 3x + 2y + 1 = 0 is
(–3/2). Let us locate the point on the curve at which the tangent is parallel to given line.

Differentiating the curve both sides with respect to x we get, 6x – 8y $\frac{dy}{dx}$ = 0

⇒ $\left( \frac{dy}{dx} \right)_{(x_{1},y_{1})}$= $\frac{3x_{1}}{4y_{1}}$ = $\frac{- 3}{2}$ [since parallel to 3x + 2y = 1]

also the point (x₁, y₁) lies on, 3x² – 4y² = 72

⇒  3$x_{1}^{2}$– 4$y_{1}^{2}$ = 72

⇒ 3$\frac{x_{1}^{2}}{y_{1}^{2}}$ – 4 = $\frac{72}{y_{1}^{2}}$

⇒ 3(4) – 4 = $\frac{72}{y_{1}^{2}}$ [as $\frac{x_{1}}{y_{1}}$ = –2]

⇒ $y_{1}^{2}$= 9

⇒ y₁ = ± 3

Required points are (–6, 3) and (6, –3)

Distance of (–6, 3) from the given line,

= $\frac{|–18 + 6 + 1|}{\sqrt{13}}$ = $\frac{11}{\sqrt{13}}$

and distance of (6, –3) from the given line,

= $\frac{|18 - 6 + 1|}{\sqrt{13}}$ = $\frac{13}{\sqrt{13}}$ =$\sqrt{13}$

Thus, (–6, 3) is the required point.