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Question : 56 of 160
Marks:
+1,
-0
Solution:
The tangent to parabola,
y2=4x at
(t2,2t) is,
y⋅2t‌‌=2(x+t2) yt‌‌=x+t2 y‌‌=‌+t‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) Normal to the ellipse,
4x2+5y2=20 or
‌+‌=1‌ at ‌(√5‌cos‌θ,2‌sin‌θ) So, the slope of normal is,
‌=‌‌tan‌θ So, the equation of the line is,
y−2‌sin‌θ‌‌=‌‌tan‌θ(x−√5‌cos‌θ) y‌‌=‌‌tan‌θ×x−‌‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) Compare (I) and (II),
t‌‌=−‌ sin‌θ‌‌=−2y‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(III) And
‌‌‌=‌‌tan‌θ sin‌θ‌‌=‌‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(IV) From equation (III) and (IV),
2t‌‌=‌ t2(4+5t2)‌‌=1 5t4+4t4‌‌=1
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