Q.
Intersection point of the tangents drawn to the parabola y2=4ax from the points (at12,2at1) and (at22,2at2) is
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Rajasthan PETRajasthan PET 2002
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Solution:
Equation of tangent to the parabola y2=4ax at the point (at12,2at1)
is 2at1=2a(x+at12) ⇒yt1=x+at12 ...(i)
Similarly, equation of tangent at the point (at22,2at2) is 2yat2=2a(x+at22) ⇒yt2=x+at22 ...(ii)
On solving Eqs. (i) and (ii), we get y=a(t1+t2),x=at1t2 ∴ Intersection point [at1t2,a(t1+t2)]