Question:
Suppose a parabola passes through \((0,4)\), \((1,9)\) and \((4,5)\) and has its axis parallel to the \(y\)-axis. Then the equation of the parabola is
Suppose a parabola passes through \((0,4)\), \((1,9)\) and \((4,5)\) and has its axis parallel to the \(y\)-axis. Then the equation of the parabola is
Show Hint
If the axis of a parabola is parallel to the \(y\)-axis, always use the form \(y=ax^2+bx+c\).
Updated On: Jun 22, 2026
- \(19x^2+12y-79x-48=0\)
- \(19x^2+12y-79x+48=0\)
- \(19y^2+12x-79y-48=0\)
- \(19y^2+12x-79y+48=0\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Write the general equation of the parabola.
Since the axis of the parabola is parallel to the \(y\)-axis, its equation is of the form
\[ y=ax^2+bx+c \]
Step 2: Use the point \((0,4)\).
Substituting \((0,4)\) into the equation,
\[ 4=a(0)^2+b(0)+c \] \[ c=4 \]
Step 3: Use the point \((1,9)\).
Substituting \((1,9)\),
\[ 9=a(1)^2+b(1)+4 \] \[ 9=a+b+4 \] \[ a+b=5 \]
Step 4: Use the point \((4,5)\).
Substituting \((4,5)\),
\[ 5=a(4)^2+b(4)+4 \] \[ 5=16a+4b+4 \] \[ 16a+4b=1 \]
Step 5: Solve for \(a\) and \(b\).
From
\[ a+b=5 \] we get
\[ b=5-a \] Substitute into
\[ 16a+4b=1 \] \[ 16a+4(5-a)=1 \] \[ 16a+20-4a=1 \] \[ 12a=-19 \] \[ a=-\frac{19}{12} \] Now,
\[ b=5+\frac{19}{12} \] \[ b=\frac{79}{12} \]
Step 6: Write the equation of the parabola.
Substitute the values of \(a\), \(b\), and \(c\):
\[ y=-\frac{19}{12}x^2+\frac{79}{12}x+4 \] Multiply throughout by \(12\):
\[ 12y=-19x^2+79x+48 \] Rearranging,
\[ 19x^2+12y-79x-48=0 \]
Step 7: Final conclusion.
Hence, the equation of the parabola is
\[ \boxed{19x^2+12y-79x-48=0} \]
Since the axis of the parabola is parallel to the \(y\)-axis, its equation is of the form
\[ y=ax^2+bx+c \]
Step 2: Use the point \((0,4)\).
Substituting \((0,4)\) into the equation,
\[ 4=a(0)^2+b(0)+c \] \[ c=4 \]
Step 3: Use the point \((1,9)\).
Substituting \((1,9)\),
\[ 9=a(1)^2+b(1)+4 \] \[ 9=a+b+4 \] \[ a+b=5 \]
Step 4: Use the point \((4,5)\).
Substituting \((4,5)\),
\[ 5=a(4)^2+b(4)+4 \] \[ 5=16a+4b+4 \] \[ 16a+4b=1 \]
Step 5: Solve for \(a\) and \(b\).
From
\[ a+b=5 \] we get
\[ b=5-a \] Substitute into
\[ 16a+4b=1 \] \[ 16a+4(5-a)=1 \] \[ 16a+20-4a=1 \] \[ 12a=-19 \] \[ a=-\frac{19}{12} \] Now,
\[ b=5+\frac{19}{12} \] \[ b=\frac{79}{12} \]
Step 6: Write the equation of the parabola.
Substitute the values of \(a\), \(b\), and \(c\):
\[ y=-\frac{19}{12}x^2+\frac{79}{12}x+4 \] Multiply throughout by \(12\):
\[ 12y=-19x^2+79x+48 \] Rearranging,
\[ 19x^2+12y-79x-48=0 \]
Step 7: Final conclusion.
Hence, the equation of the parabola is
\[ \boxed{19x^2+12y-79x-48=0} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Parabola Questions
- Find the distance between the point \( (6, 4\sqrt{3}) \) and the focus of the parabola \( y^2 = 8x \)
- The equation of the normal to the parabola \[ y^2=12x \] at the point \((3\lambda^2,6\lambda)\) is
- If \[ y=mx+\frac{3}{m} \] is a tangent to the parabola \[ y^2=4ax \] at the point \(P(3,\beta)\), where \(\beta<0\), then the value of \[ 3m-\beta \] is:
- The product of the slopes of the non-horizontal normals drawn through the point \[ (6,0) \] to the parabola \[ y^2=8x \] is:
- Which of the following represents a parabola?
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions