Question

If equation of the parabola is $(6x+8y-5{)}^2=60(8x-6y+9),$ then

• A. Equation of axis of the parabola is $6x+8y-5=0$
• B. Equation of tangent at vertex of the parabola is $6x+8y-5=0$
• C. Length of latus rectum is $\frac{3}{2}$ units
• D. Length of latus rectum is $6$ units

Answer 1

Answer: (A), (D)

Length of latus rectum is $6$ units

$(6x+8y-5{)}^2=60(8x-6y+9)$
$\Rightarrow {\left(\frac{6x+8y-5}{\sqrt{6^2+8^2}}\right)}^2=6\times \left(\frac{8x-6y+9}{\sqrt{8^2+6^2}}\right)$

On comparing with ${\text{(Distance from axis)}}^2=L.L.R.\text{(distance from T.V.)},$
equation of axis of the parabola is $6x+8y-5=0$
equation of tangent at vertex of the parabola is $8x-6y+9=0$
Length of latus rectum $(L.L.R.)=6$ units