TS EAMCET · Maths · Parabola
If a normal to the parabola \(y^2=12 x\) at \(A(3,-6)\) cuts the parabola again at \(P\), then the equation of the tangent at \(P\) is
- A \(x-3 y+27=0\)
- B \(x+y=45\)
- C \(y-x+9=0\)
- D \(3 x+y=99\)
Answer & Solution
Correct Answer
(B) \(x+y=45\)
Step-by-step Solution
Detailed explanation
We have, \(y^2=12 x\) \(\therefore \quad 2 y \frac{d y}{d x}=12 \Rightarrow \frac{d y}{d x}=\frac{6}{y}\) \(\therefore\) Slope of Normal at \(A(3,-6)=\frac{-1}{\left.\frac{d y}{d x}\right|_{(3,-6)}}=\frac{-1}{\left(\frac{6}{-6}\right)}=1\) \(\therefore\) Equation of normal at…
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
- If \(\mathbf{e}\) is a unit vector perpendicular to the plane determined by the points \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\). If \(\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\), then the projection vector of \(\mathbf{a}\) on \(\mathbf{e}\) isTS EAMCET 2018 Medium
- For all real values of \(\mathrm{x}\), the minimum value of \(\frac{1-x+x^2}{1+x+x^2}\) isTS EAMCET 2023 Medium
- The range of ' \(a\) ' so that \(a^2 x^2+2 x y+4 y^2=0\) represents two distinct lines isTS EAMCET 2023 Easy
- The direction cosines of the line making angles and respectively with and axes, areTS EAMCET 2021 Medium
- The area enclosed by the curves \(y=8 x-x^2\) and \(8 x-4 y+11=0\) isTS EAMCET 2018 Medium
- Let \(\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) and \(\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\). If \(\mathbf{p}\) is a unit vector such that \([\mathbf{a b p}]\) is maximum. then \(\mathbf{p}=\)TS EAMCET 2019 Easy
More PYQs from TS EAMCET
- Commercially available \(\mathrm{H}_2 \mathrm{SO}_4\) is \(98 \mathrm{~g}\) by weight of \(\mathrm{H}_2 \mathrm{SO}_4\) and \(2 \mathrm{~g}\) by weight of water. It's density is \(1.38 \mathrm{~g} \mathrm{~cm}^{-3}\). Calculate the molality \((m)\) of \(\mathrm{H}_2 \mathrm{SO}_4\) (molar mass of \(\mathrm{H}_2 \mathrm{SO}_4\) is \(98 \mathrm{~g} \mathrm{~mol}^{-1}\) )TS EAMCET 2017 Medium
- Two objects are located at height above the ground. At some point of time, the objects are thrown with initial velocity at an angle and with the positive -axis respectively. Assuming , the velocity vectors will be perpendicular to each other at time equal toTS EAMCET 2018 Medium
- \(y=A e^x+B e^{2 x}+C e^{3 x}\) satisfies the differential equationTS EAMCET 2004 Medium
- Suppose \(\alpha, \beta, \gamma\) are the roots of \(x^3+x^2+x+2=0\). Then, the value of \(\left(\frac{\alpha+\beta-2 \gamma}{\gamma}\right)\left(\frac{\beta+\gamma-2 \alpha}{\alpha}\right)\left(\frac{\gamma+\alpha-2 \beta}{\beta}\right)\) isTS EAMCET 2015 Easy
- The heat required to convert 8 g of ice at a temperature of \(-20^{\circ} \mathrm{C}\) to steam at \(100^{\circ} \mathrm{C}\) is [Specific heat capacity of ice \(=2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\), specific heat capacity of water \(=4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\), latent heat of fusion of ice \(=336 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1}\) and latent heat of steam \(\left.=2.268 \times 10^6 \mathrm{Jkg}^{-1}\right]\)TS EAMCET 2025 Medium
- By using which process, sodium carbonate is generally prepared?TS EAMCET 2025 Easy