Question:
If the focus of the parabola
\[
(y - k)^2 = 4(x - h)
\]
always lies between the lines \(x + y = 1\) and \(x + y = 3\) then:
If the focus of the parabola
\[
(y - k)^2 = 4(x - h)
\]
always lies between the lines \(x + y = 1\) and \(x + y = 3\) then:
Show Hint
Understanding how a parabola’s focus is derived from its equation is key to solving locus-related problems.
Updated On: Feb 3, 2025
- \(0<h + k<2\)
- \(0<h + k<1\)
- \(1<h + k<2\)
- \(1<h + k<3\)
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
The standard form of the given parabola is:
\[
(y - k)^2 = 4(x - h)
\]
The focus of the parabola is given by:
\[
(h + 1, k)
\]
Since the focus must lie between the lines \(x + y = 1\) and \(x + y = 3\), we substitute the focus into the inequalities:
\[
1<(h+1) + k<3
\]
\[
0<h + k<2
\]
Thus, the required range for \( h + k \) is:
\[
0<h + k<2
\]
Was this answer helpful?
0
0
Top Questions on Parabola
- Let \( P(4, 4\sqrt{3}) \) be a point on the parabola \( y^2 = 4ax \) and PQ be a focal chord of the parabola. If M and N are the foot of the perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:
- Let \(L_1\) be the length of the common chord of the curves \[ x^2 + y^2 = 9 \quad {and} \quad y^2 = 8x \] and let \(L_2\) be the length of the latus rectum of \(y^2 = 8x\). Then:
- Find the area of the region bounded by the parabola \[ y^2 = 4ax \text{ and its latus rectum.} \]
- The equation of the parabola whose focus is $(6, 0)$ and directrix is $x = -6$ is:
- Let \( L_1, L_2 \) be the lines passing through the point \( P(0, 1) \) and touching the parabola \[ 9x^2 + 12x + 18y - 14 = 0. \] Let \( Q \) and \( R \) be the points on the lines \( L_1 \) and \( L_2 \) such that the \( \Delta PQR \) is an isosceles triangle with base \( QR \). If the slopes of the lines \( QR \) are \( m_1 \) and \( m_2 \), then \( 16\left(m_1^2 + m_2^2\right) \) is equal to ______.
View More Questions
Questions Asked in BITSAT exam
- For two events A and B, if \(P(A) = P(A/B) = \frac{1}{4}\) and \(P(B/A) = \frac{1}{2}\), then which of the following is not true?
- Given below is the distribution of a random variable \(X\):
\[ \begin{array}{|c|c|} \hline X = x & P(X = x) \\ \hline 1 & \lambda \\ 2 & 2\lambda \\ 3 & 3\lambda \\ \hline \end{array} \] If \(\alpha = P(X<3)\) and \(\beta = P(X>2)\), then \(\alpha : \beta = \)- BITSAT - 2024
- Probability
- The probability of getting 10 in a single throw of three fair dice is:
- BITSAT - 2024
- Probability
- Let the foot of perpendicular from a point \( P(1,2,-1) \) to the straight line \( L : \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) be \( N \). Let a line be drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \) which meets \( L \) at point \( Q \). If \( \alpha \) is the acute angle between the lines \( PN \) and \( PQ \), then \( \cos \alpha \) is equal to:
- If \( \vec{a} = 2\hat{i} + \hat{j} + 2\hat{k} \), then the value of \( |\hat{i} \times (\vec{a} \times \hat{i})| + |\hat{j} \times (\vec{a} \times \hat{j})| + |\hat{k} \times (\vec{a} \times \hat{k})|^2 \) is equal to:}
- BITSAT - 2024
- Algebra
View More Questions