The parabola y 2 = ax cuts x 2 –y 2 = 2a 2 at P and Q. Tangent at P to the hyperbola cuts the parabola again in R. Find the area of the curvilinear triangle PQR.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. Solving y 2 = ax and x 2 –y 2 = 2a 2 together, we obtain
the coordinates of P and Q as
(2a, a 
) and (2a, –a 
) respectively. The equation of the tangent to x 2 –y 2 = 2a 2 at P (2a,a 
) is 2ax –a 
y = 2a 2 or, 2x – 
y = 2a
Solving this equation with y 2 = ax, we obtain the coordinates fo R as (a/2, –a/ 
)
Now,
Area of curve linear Δ PQR
= Area of curvilinear Δ QRS + Area of curvilinear triangle PRS.
= +
= 
+
= 
+ 
– 
– 
= 
– 
– 
=2 
– 
– 
= 
– 
– 
= [2 
a 2 + 2a 2 log (a 
+ 2a) –2a 2 log ( 
a)]
– 
– 
= 
= 
sq. units.
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