AP EAMCET · Maths · Parabola
If the focus of a parabola is \((0,-3)\) and its directrix is \(y=3\), then it equation is
- A \(x^2=12 y\)
- B \(y^2=-12 x\)
- C \(y^2=12 x\)
- D \(x^2=-12 y\)
Answer & Solution
Correct Answer
(D) \(x^2=-12 y\)
Step-by-step Solution
Detailed explanation
This is of the form of \(x^2=-4 a y\) Here, \(S \equiv(0,-a) \equiv(0,-3)\) Thus, \(a=3\) \(\therefore\) Equation of parabola is \(x^2=-4 \times 3 y\) \(\Rightarrow \quad x^2=-12 y\)
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 the angles of a triangle are in the ratio \(1: 2: 3\), the corresponding sides are in the ratioAP EAMCET 2021 Easy
- The normal to the curve \(y=f(x)\) at the points \((3,4)\) makes an angle \(\frac{3 \pi}{4}\) with positive \(X\)-axis then \(f^{\prime}(3)=\)AP EAMCET 2022 Easy
- If \(0 < p < q\), then \(\lim _{n \rightarrow \infty}\left(q^n+p^n\right)^{1 / n}\) is equal to :AP EAMCET 2006 Medium
- \(\int \frac{\log _e x}{\left(1+\log _e x\right)^2} d x=\)AP EAMCET 2017 Medium
- The angle between the tangents drawn from a point \((-3,2)\) to the ellipse \(4 x^2+9 y^2-36=0\) isAP EAMCET 2025 Medium
- If a line makes angles \(\frac{\pi}{4}\) and \(\frac{\pi}{3}\) with \(Y\)-axis and \(Z\)-axis respectively, then the obtuse angle made by that line with \(X\)-axis isAP EAMCET 2018 Easy
More PYQs from AP EAMCET
- Suppose the pairs of straight lines \(x^2-2 a x y-y^2=0\) and \(x^2-2 b x y-y^2=0\) are such that each pair bisects the angles between the other two. Then \(a b=\)AP EAMCET 2022 Medium
- If the sides \(a, b, c\) of the triangle \(A B C\) are in harmonic progression, then \(\operatorname{cosec}^2 \mathrm{~A} / 2, \operatorname{cosec}^2 \mathrm{~B} / 2, \operatorname{cosec}^2 \mathrm{C} / 2\) are inAP EAMCET 2025 Medium
- In \(\triangle A B C\), if \(a=5, b=4\) and \(\cos (A-B)=\frac{31}{32}\), then \(c=\)AP EAMCET 2025 Medium
- At \(\mathrm{T}(\mathrm{K})\), the vapour pressure of pure benzene (molar mass \(=78 \mathrm{~g} \mathrm{~mol}^{-1}\) ) is \(0.85 \mathrm{bar}\). When \(2.0 \mathrm{~g}\) of non-volatile, non-electrolyte solute is added to \(39 \mathrm{~g}\) of benzene, the vapour pressure of solution at \(\mathrm{T}(\mathrm{K})\) is 0.83 bar. The elevation in boiling point (in \(\mathrm{K}\) ) of the same solution is: \(\left(\mathrm{K}_{\mathrm{b}}\right.\) of benzene is \(2.6 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\))AP EAMCET 2017 Medium
- A point \(C\) with position vector \(\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3}\) (where \(\bar{a}, \bar{b}\) and \(\bar{c}\) are non coplanar vectors) divides the line joining \(A\) and \(B\) in the ratio \(2: 1\). If the position vector of \(A\) is \(\bar{a}-2 \bar{b}+3 \bar{c}\), then the position vector of \(B\) isAP EAMCET 2017 Easy
- If \(\alpha, \beta, \gamma\) are the roots of \(x^3+2 x+5=0\), then \(\sum \frac{\beta+\gamma}{\alpha^2}=\)AP EAMCET 2023 Medium