If $(h, k)$ is the point to which the origin is shifted in order to transform the equation $y^2-4x+6y+17=0$ into the form $Y^2=4aX$,then $h^2+k^2=$
- A$11$
- B$1$
- C$25$
- D$13$
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The line $y = 2x + c$ is a tangent to the parabola $y^2 = 16x$,if $c$ equals
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View SolutionConsider the parabola $25[(x-2)^2+(y+5)^2]=(3x+4y-1)^2$. Match the characteristics of this parabola given in List-$I$ with their corresponding items in List-$II$.
List-$I$ List-$II$ $I$. Vertex $A$. $8$ $II$. Length of latus rectum $B$. $(\frac{29}{10}, \frac{-38}{10})$ $III$. Directrix $C$. $3x+4y-1=0$ $IV$. One end of the latus rectum $D$. $(\frac{-2}{5}, \frac{-16}{5})$ $E$. $6$
| List-$I$ | List-$II$ |
| $I$. Vertex | $A$. $8$ |
| $II$. Length of latus rectum | $B$. $(\frac{29}{10}, \frac{-38}{10})$ |
| $III$. Directrix | $C$. $3x+4y-1=0$ |
| $IV$. One end of the latus rectum | $D$. $(\frac{-2}{5}, \frac{-16}{5})$ |
| $E$. $6$ |
The equation of the tangent to the parabola $y^2=12x$ at the point $(3,-6)$ is:
If the vertex of a parabola is $(4,3)$ and its directrix is $3x+2y-7=0$,then the equation of the latus rectum of the parabola is
The locus of the vertices of the family of parabolas $6y = 2a^3x^2 + 3a^2x - 12a$ is
MediumWBJEE 2021
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