JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(r\) be the radius of the circle, which touches x -axis at point \((\mathrm{a}, 0), \mathrm{a} \lt 0\) and the parabola \(\mathrm{y}^2=9 \mathrm{x}\) at the point \((4,6)\). Then \(r\) is equal to ________
- A 10
- B 20
- C 30
- D 40
Answer & Solution
Correct Answer
(C) 30
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & (x-a)^2+(y-r)^2=r^2 \\ & (4-a)^2+(6-r)^2=r^2 \\ & 16+a^2-8 a+36+r^2-12 r=r^2 \\ & a^2-8 a-12 r+52=0 \end{aligned}\) Tangent to parabola at \((4,6)\) is \(6.4=9 .\left(\frac{x+4}{2}\right) \text { i.e. } 3 x-4 y+12=0\) This is also tangent to the circle…
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