Question

If the line ax + by = 1 passes through point of intersection of y = x tana + p seca, y sin(30⁰ – a) – x cos(30⁰ – a) = p and is inclined at 30⁰ with y = tanax, then the value of a² + b² can be-

• A. $\frac{1}{p^{2}}$
• B. $\frac{2}{p^{2}}$
• C. $\frac{3}{2p^{2}}$
• D. $\frac{3}{4p^{2}}$

Answer 1

Answer: (D)

$\frac{3}{4p^{2}}$

Answer 2

Given, y cosa – x sina = p

and ysin(30⁰ – a) – x cos(30⁰ – a) = p

are inclined at 60⁰ so line ax + by = 1 can be acute angle bisector ....(i)

i.e., y cosa – x sina – p

= –(ysin(30⁰ – a) – xcos(30⁰ – a) – p)

Ю y[cosa+sin(30⁰–a)]

–x[sina + cos(30⁰–a)] = 2p ....(ii)

From Eqs.(i) and (ii), we get

$\frac{b}{\cos\alpha + \sin(30º - \alpha)}$ = $\frac{a}{(\sin\alpha + \cos(30º - \alpha))}$ = $\frac{1}{2p}$

Ю $\frac{\sqrt{a^{2} + b^{2}}}{\sqrt{2 + 1}}$ = $\frac{1}{2p}$

Ю a² + b² = $\frac{3}{4p^{2}}$