Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If the area of the triangle whose one vertex is at the vertex of the parabola, $y^{2}+4(x-a^{2})=0$ and the other two vertices are

the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :
Option 1)
$5\sqrt5$
Option 2)
$5(2^{\frac{1}{3}})$
Option 3)
$(10)^{\frac{2}{3}}$
Option 4)
5

Answers (1)

best_answer

Standard equation of parabola -

$y^{2}=4ax$

- wherein

Parametric coordinates of parabola -

$x= at^{2}$

$y= 2at$

- wherein

For the parabola.

$y^{2}=4ax$

Vertex is $(a^{2},0)$

$y^{2}=-4(x-a^{2})$

Parabola intersect y-axis (x = 0)

$=>y^{2}=-4(0-a^{2})$

$=>y=\pm 2a$

Area of triangle = $\frac{1}{2}\cdot 4a\cdot a^{2}=250$

$=>a^{3}=125$

$=>a=5$

Option 1)

$5\sqrt5$

Option 2)

$5(2^{\frac{1}{3}})$

Option 3)

$(10)^{\frac{2}{3}}$

Option 4)
5

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year

anam-khan

What after JEE Main results 2024? All you need to know about JEE Adv, CSAB, JoSAA counselling

Latest Question