JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the tangent to the conic, \(y - 6 = x^2\) at \((2, 10)\) touches the circle, \(x^2 + y^2 + 8x - 2y = k\) (for some fixed \(k\) ) at a point \((\alpha ,\,\beta );\) then \((\alpha ,\,\beta )\) is
- A \(\left( { - \frac{7}{{17}},\frac{6}{{17}}} \right)\)
- B \(\left( { - \frac{4}{{17}},\frac{1}{{17}}} \right)\)
- C \(\left( { - \frac{6}{{17}},\frac{10}{{17}}} \right)\)
- D \(\left( { - \frac{8}{{17}},\frac{2}{{17}}} \right)\)
Answer & Solution
Correct Answer
(D) \(\left( { - \frac{8}{{17}},\frac{2}{{17}}} \right)\)
Step-by-step Solution
Detailed explanation
\({x^2}y + 6 = 0\) \(2x - \frac{{dy}}{{dx}} = 0 \Rightarrow \frac{{dy}}{{dx}} = 2x\) \({\left. {\frac{{dy}}{{dx}}} \right|_{\left( {x,y} \right) = \left( {2,10} \right)}} = 4\) Equation of tangent \(y-10=4(x-z)\) \(4x-y+z=0\) tangent passes through…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- \(\lim \limits_{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3 a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}}(a \neq 0)\) is equal toJEE Mains 2020 Hard
- Let \(Q\) be the foot of the perpendicular from the point \(\mathrm{P}(7,-2,13)\) on the plane containing the lines \(\frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8}\) and \(\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}\) Then \((\mathrm{PQ})^{2}\), is equal to ..... .JEE Mains 2021 Hard
- If the tangent at a point on the ellipse \(\frac{{{x^2}}}{{27}} + \frac{{{y^2}}}{3} = 1\) meets the coordinate axes at \(A\) and \(B,\) and \(O\) is the origin, then the minimum area (in sq. units) of the triangle \(OAB\) isJEE Mains 2016 Hard
- The area of the region bounded by the curve \(y=\max \{|x|, x|x-2|\}\), then \(x\)-axis and the lines \(x=-2\) and \(x=4\) is equal to _______ .JEE Mains 2025 Easy
- Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be a solution of the differential equation, \(\sqrt{1-\mathrm{x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}+\sqrt{1-\mathrm{y}^{2}}=0,|\mathrm{x}|<1\) If \(\mathrm{y}\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2},\) then \(\mathrm{y}\left(\frac{-1}{\sqrt{2}}\right)\) is equal toJEE Mains 2020 Hard
- Given below are two statements :
Statement I : \(\lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x+\log _e \sqrt{\frac{1+x}{1-x}}-2 x}{x^5}\right)=\frac{2}{5}\)
Statement II : \(\lim _{\mathrm{x} \rightarrow 1}\left(\mathrm{x}^{\frac{2}{1-\mathrm{x}}}\right)=\frac{1}{\mathrm{e}^2}\)
In the light of the above statements, choose the correct answer from the options given below :JEE Mains 2025 Medium
More PYQs from JEE Mains
- Let \(\vec{a}\) and \(\vec{b}\) be the vectors along the diagonal of a parallelogram having area \(2 \sqrt{2}\). Let the angle between \(\vec{a}\) and \(\vec{b}\) be acute. \(|\vec{a}|=1\) and \(|\vec{a} . \vec{b}|=|\vec{a} \times \vec{b}| .\) If \(\vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}\), then an angle between \(\vec{b}\) and \(\vec{c}\) isJEE Mains 2022 Hard
- Let \(y=y(x)\) be the solution of the differential equation \(x\sqrt{1-x^2}\,dy + \left(y\sqrt{1-x^2} - x\cos^{-1}x\right)dx = 0\), \(x \in (0, 1)\), \(\displaystyle\lim_{x\to 1^-} y(x) = 1\). Then \(y\left(\dfrac{1}{2}\right)\) equals:JEE Mains 2026 Medium
- The value of the integral \(\displaystyle\int_{\pi/6}^{\pi/3} \left(\dfrac{4 - \csc^2 x}{\cos^4 x}\right) dx\) is:JEE Mains 2026 Hard
- The distance of the line \(\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}\) from the point \((1,4,0)\) along the line \(\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}\) is :JEE Mains 2025 Hard
- Let \(A\) be a \(3 \times 3\) invertible matrix. If \(|adj (24 A ) \mid=\) \(\operatorname{adj}(3 \operatorname{adj}(2 A )) \mid\), then \(\mid A ^{2}|=\dots\dots\dots\) is equal toJEE Mains 2022 Hard
- The probability of a man hitting a target is \(\frac{1}{10}\). The least number of shots required, so that the probability of his hitting the target at least once is greater than \(\frac{1}{4},\) isJEE Mains 2020 Hard