Solveeit Logo
Login

Question

Question: Equation of parabola having the extremities of it’s latus rectum as (3, 4) and (4, 3) is –...

Equation of parabola having the extremities of it’s latus rectum as (3, 4) and (4, 3) is –

A

(x−72)2\left( x - \frac{7}{2} \right)^{2}+ (y−72)2\left( y - \frac{7}{2} \right)^{2}= (x+y−62)2\left( \frac{x + y - 6}{2} \right)^{2}

B

(x−72)2\left( x - \frac{7}{2} \right)^{2}+ (y−72)2\left( y - \frac{7}{2} \right)^{2}= (x+y−8)22\frac{(x + y - 8)^{2}}{2}

C

(x−72)2\left( x - \frac{7}{2} \right)^{2}+ (y−72)2\left( y - \frac{7}{2} \right)^{2}= (x+y−4)22\frac{(x + y - 4)^{2}}{2}

D

None of these

Answer

(x−72)2\left( x - \frac{7}{2} \right)^{2}+ (y−72)2\left( y - \frac{7}{2} \right)^{2}= (x+y−8)22\frac{(x + y - 8)^{2}}{2}

Explanation

Solution

Focus is (72,72)\left( \frac{7}{2},\frac{7}{2} \right)and it’s axis is the line y = x.

corresponding value of ‘a’ is 14\frac{1}{4} (1+1\sqrt{1 + 1}) = 24\frac{\sqrt{2}}{4}. Let the equation of its directrix be y + x + l = 0

Ž ∣3+4+λ∣2\frac{|3 + 4 + \lambda|}{\sqrt{2}} = 2. 24\frac{\sqrt{2}}{4} Ž l = –6, –8.

Thus equation of parabola is

(x−72)2+(y−72)2\left( x - \frac{7}{2} \right)^{2} + \left( y - \frac{7}{2} \right)^{2}=(x+y−6)22\frac{(x + y - 6)^{2}}{2}

(x−72)2+(y−72)2\left( x - \frac{7}{2} \right)^{2} + \left( y - \frac{7}{2} \right)^{2}= (x+y−8)22\frac{(x + y - 8)^{2}}{2}.