Conic Sections Question 181
Question: Tangent to the parabola $ y=x^{2}+6 $ at (1, 7) touches the circle $ x^{2}+y^{2}+16x+12y+c=0 $ at the point
[IIT Screening 2005]
Options:
A) (-6, -9)
B) (-13, -9)
C) (-6, -7)
D) (13, 7)
Show Answer
Answer:
Correct Answer: C
Solution:
Equation of tangent at (1, 7) to $ y=x^{2}+6 $
$ \frac{1}{2}(y+7)=x.1+6 $
therefore $ y=2x+5 $ ……(i) This tangent also touches the circle $ x^{2}+y^{2}+16x+12y+c=0 $
…..(ii) Now solving (i) and (ii), we get
therefore $ x^{2}+{{(2x+5)}^{2}}+16x+12(2x+5)+c=0 $
therefore $ 5x^{2}+60x+85+c=0 $
Since, roots are equal so $ b^{2}-4ac=0 $
therefore $ {{(60)}^{2}}-4\times S\times (85+c)=0 $
therefore $ 85+c=180 $
therefore $ 5x^{2}+60x+180=0 $
therefore $ x=-\frac{60}{10}=-6 $
therefore $ y=-7 $
Hence, point of contact is (-6, -7).
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