Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

area of triangle formed by pair of lines (x+2a)^2-3y^2 and x=a

area of triangle formed by pair of lines (x+2a)^2-3y^2 and x=a

Grade:11

1 Answers

Ramvardhan
25 Points
6 years ago
The pair of lines (x+2a)2 – 3y2 = 0 passes through the origin ( just substitiute 0,0 and if it satisfies then it passes through the origin ). Now this pair of lines along with x=a forms a triangle. x=a must be a line parallel to the x-axis either in the left or right side. The pair of lines is in the form of an “X” with its pt. of intersection at the origin. So, (0,0) is one of the vertex. x=a and (x+2a)2 – 3y2 = 0 intersect at 2 points.
Solving botht the equations:
Substitute x=a in (x+2a)2 – 3y2 = 0 as on the pt. of intersection, that pt. should satisfy botht the lines.
So you will get,
(3a)2 – 3y2 = 0.
So, 3a2 = y2
So y = a3, -a3
So the points of intersection are (a,a3) and (a, -a3).
So now we know all the 3 vertices of the triangle : (0,0), (a,a3) and (a, -a3).
 
Now area of the triangle is : 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]
Substitute the vertices in this equation and find the area. The final answer is 3a2
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free