If the vertex of the conic y^{2}-4y=4x-4a alw | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equation of the conic is $y^{2}-4y=4x-4a$.Completing the square for $y$, we get $y^{2}-4y+4=4x-4a+4$, which simplifies to $(y-2)^{2}=4(x

Vedclass · Accepted Answer

$-\frac{1}{2} < a < 2$

Vedclass · Answer

(B) The given equation of the conic is $y^{2}-4y=4x-4a$.Completing the square for $y$, we get $y^{2}-4y+4=4x-4a+4$, which simplifies to $(y-2)^{2}=4(x-a+1)$.Comparing this with the standard form $(y-k)^{2}=4A(x-h)$, the vertex is $(h, k) = (a-1, 2)$.Let the vertex be $(x_{0}, y_{0}) = (a-1, 2)$.The vertex lies between the lines $x+y-3=0$ and $2x+2y-1=0$.Rewriting the lines in terms of $x+y$:Line $1$: $x+y=3$.Line $2$: $2(x+y)=1 \implies x+y=\frac{1}{2}$.Since the vertex lies between these two parallel lines, the value of $x+y$ for the vertex must lie between the values of $x+y$ for the lines.Thus, $\frac{1}{2} Substituting $x_{0}+y_{0} = (a-1)+2 = a+1$:$\frac{1}{2} Subtracting $1$ from all parts:$\frac{1}{2}-1 < a < 3-1$, which gives $-\frac{1}{2} < a < 2$.

If the vertex of the conic $y^{2}-4y=4x-4a$ always lies between the straight lines $x+y=3$ and $2x+2y-1=0$, then

Similar Questions

The solution set of $(5+4 \cos \theta)(2 \cos \theta+1)=0$ in the interval $[0, 2 \pi]$ is:

$A$ radioactive nucleus (initial mass number $A$ and atomic number $Z$) emits $3$ $\alpha$-particles and $2$ positrons. The ratio of neutrons to protons in the final nucleus will be:

$A$ force $F$ is applied on a square area of side $L$. If the percentage error in the measurement of $L$ is $2 \%$ and that in $F$ is $4 \%$,what is the maximum percentage error in pressure?

Which one of the following pairs of organisms are exotic species introduced in India?

The output $(X)$ of the logic circuit shown in the figure will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module