Question

A common tangent to 9x² – 16y² = 144 and x² + y² = 9 is –

• A. y = $\frac{3x}{\sqrt{7}}$ + $\frac{15}{\sqrt{7}}$
• B. y = 3$\sqrt{\frac{2}{7}}$x + $\frac{15}{\sqrt{7}}$
• C. y = 2$\sqrt{\frac{3}{7}}$x + $\frac{15}{\sqrt{7}}$
• D. None of these

Answer 1

Answer: (B)

y = 3$\sqrt{\frac{2}{7}}$x + $\frac{15}{\sqrt{7}}$

Answer 2

Let y = mx + c be a common tangent to

9x² – 16y² = 144 and x² + y² = 9.

Since, y = mx + c is a tangent to 9x² – 16y² = 144, therefore,

c² = a²m² – b² Ž c² = 16m² – 9 … (1)

Now, y = mx + c is a tangent to x² + y² = 9, therefore,

$\frac{c}{\sqrt{m^{2} + 1}}$ = 3 Ž c² = 9(1 + m²) … (2)

From equations (1) and (2), we get

16m² – 9 = 9 + 9m² Ž m = ± 3$\sqrt{\frac{2}{7}}$

On putting the value of m in equation (2), we get

c = ± $\frac{15}{\sqrt{7}}$

Hence, y = 3 $\sqrt{\frac{2}{7}}$x + $\frac{15}{\sqrt{7}}$ is a common tangent.