Question

The locus of mid point of that chord of parabola which subtends right angle on the vertex will be

• A. y² – 2ax + 8a² = 0
• B. y² = a(x – 4a)
• C. y² = 4a(x – 4a)
• D. y² + 3ax+4a² = 0

Answer 1

Answer: (A)

y² – 2ax + 8a² = 0

Answer 2

Equation of parabola y² = 4ax

Equation of that chord of parabola whose mid point is (x₁, y₁) will be yy₁ = 2a(x+x₁) = $y_{1}^{2} - 4ax_{1}$ or yy₁ – 2ax

= $y_{1}^{2} - 2ax_{1}$ or $\frac{yy_{1} - 2ax}{y_{1}^{2} - 2ax_{1}} = 1$ ......... (ii)

Making equation (i) homogeneous by equation (ii), the equation of lines joining the vertex (0,0) of parabola to the point of intersection of chord (ii) and parabola (i) will be

y² = 4ax $\frac{yy_{1} - 2ax}{y_{1}^{2} - 2ax_{1}}$ of y²($y_{1}^{2} - 2ax_{1}$) = 4a(yy₁ – 2ax) or 8a²x²- 4ay₁xy + $(y_{1}^{2}$ - 2ax₁)y² = 0.

If lines represented by it are mutually perpendicular, then coefficient of x² + coefficient of y²=0. Therefore,

8a² + ($y_{1}^{2}$ - 2ax₁) = 0 or $y_{1}^{2}$ - 2ax₁ + 8a² = 0.

∴ Required locus of (x₁, y₁) is y² – 2ax + 8a² = 0.