Question

6. If the segment intercepted by the parabola \( y=4 a x \) with the line \( l x+m y+n=0 \) subtends a right angle at the vertex, then (a) \( 4 a l+n=0 \) (b) \( 4 a l+4 a m+n=0 \) (c) \( 4 a m+n=0 \) (d) \( a l+n=0 \)

Answer 1

Given parabola is y² = 4ax.

And given line is lx+my+n = 0.

According to given condition OQ \(\bot\) OP where P and Q are points at which line intersects parabola.

Let point P and Q are \(P(at^2_1,2at_1)\) and \(Q(at^2_2,2at_2)\)

Slope of line OP is m₁ \(=\frac{2at_1}{at^2_1}=\frac2{t_1}\)

And slope of line OQ is m₂ \(=\frac{2at_2}{at^2_2}=\frac2{t_2}\)

∵ OQ \(\bot\) OP

∴ m₁m₂ = -1

⇒ 2/t₁ x 2/t₂ = -1

⇒ t₁t₂ = -4  ...(1)

Since, point P and Q lies on line lx+my+n = 0.

∴  \(lat^2_1+2mat_1+n=0\) (Point P satisfies equation of line)

And \(lat^2_2+2mat_2+n=0\) (Point Q satisfies equation of line)

As t₁ and t₂ are roots of lat+2mat+n = 0.

∴ t₁t₂ = Product of roots = c/a = n/la

⇒ n/la = -4 (From equation (i))

⇒ n = -4la

⇒ n+4la = 0

Hence option (a) is correct.