120090
If the straight line \(y=m x+c\) is parallel to the axis of the parabola \(y^2=l x\) and intersects the parabola at \(\left(\frac{c^2}{8}, c\right)\), then the length of the latus rectum is
1 2
2 3
3 4
4 8
Explanation:
D Given, \(\mathrm{y}^2=l \mathrm{x},\left(\frac{\mathrm{c}^2}{8}, \mathrm{c}\right)\)
Length of latus rectum of parabola is \(4 a=l\) We can write -
\(\mathrm{c}^2=l \times \frac{\mathrm{c}^2}{8}\)
\(l=8\)
Also,
\(l=4 \mathrm{a}=8\)
AP EAMCET-2011
Parabola
120091
Let \(M\) be the foot of the perpendicular from a point \(P\) on the parabola \(y^2=8(x-3)\) onto its directrix and let \(S\) be the focus on the parabola. If \(\triangle \mathrm{SPM}\) is an equilateral triangle, then \(P\) is equal to
120092
If the double ordinate of the parabola \(y^2=8 x\) is of Length 16, then the angle subtended by it at the vertex of the parabola is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{3 \pi}{4}\)
4 \(\frac{\pi}{4}\)
Explanation:
A Given,
\(\mathrm{y}^2=8 \mathrm{x}=4\). (2) \(\mathrm{x}\)
Double ordinate \(=16\)
Let end points of double ordinate \(=\left(2 t^2 \pm 4 t\right)\)
Length \(=16=8 \mathrm{t}\)
\(\Rightarrow \mathrm{t}=2\)
Endpoint- \(\mathrm{P}(8,-8)\) \& \(\mathrm{Q}(8,8)\)
\(\mathrm{PQ}=\sqrt{(8-8)+(-8-8)^2}=16\)
\(\mathrm{OP}=8 \sqrt{2}\)
\(\mathrm{OQ}=8 \sqrt{2} \quad(\therefore \mathrm{O}\) is origin \()\)
\((\mathrm{OP})^2+(\mathrm{OQ})^2=(\mathrm{PQ})^2\)
\(\therefore\) The angle subtended by double ordinate is ' \(\frac{\pi}{2}\),
AP EAMCET-24.04.2018
Parabola
120093
If a normal chord at a point \(t\) on the parabola \(y^2=4 a x\) subtends a right angle at the vertex then \(t\) equals to
1 1
2 \(\sqrt{2}\)
3 2
4 \(\sqrt{3}\)
Explanation:
B Given,
\(y^2=4 \mathrm{ax}\)
Let, end points \(\left(a t_1^2, 2 a_1\right)\) and \(\left(a t_2^2, 2 a t_2\right)\)
\(\because\) it is normal to parabola :
\(\mathrm{t}_2=-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}\)
Equation of normal to parabola -
\(y=t x+2 a t+a t^3\)
Slope is- \(t_1\)
Now, chord is perpendicular at origin so-
\(\frac{2 \mathrm{at}_1-0}{\mathrm{at}_1^2-0}=\frac{2 \mathrm{at}_1-0}{\mathrm{at}_2^2-0}=-1\)
\(\mathrm{t}_1 \mathrm{t}_2=-4\)
\(\mathrm{t}_1\left(-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}\right)=-4\)
\(\mathrm{t}_1=\sqrt{2}\)
120090
If the straight line \(y=m x+c\) is parallel to the axis of the parabola \(y^2=l x\) and intersects the parabola at \(\left(\frac{c^2}{8}, c\right)\), then the length of the latus rectum is
1 2
2 3
3 4
4 8
Explanation:
D Given, \(\mathrm{y}^2=l \mathrm{x},\left(\frac{\mathrm{c}^2}{8}, \mathrm{c}\right)\)
Length of latus rectum of parabola is \(4 a=l\) We can write -
\(\mathrm{c}^2=l \times \frac{\mathrm{c}^2}{8}\)
\(l=8\)
Also,
\(l=4 \mathrm{a}=8\)
AP EAMCET-2011
Parabola
120091
Let \(M\) be the foot of the perpendicular from a point \(P\) on the parabola \(y^2=8(x-3)\) onto its directrix and let \(S\) be the focus on the parabola. If \(\triangle \mathrm{SPM}\) is an equilateral triangle, then \(P\) is equal to
120092
If the double ordinate of the parabola \(y^2=8 x\) is of Length 16, then the angle subtended by it at the vertex of the parabola is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{3 \pi}{4}\)
4 \(\frac{\pi}{4}\)
Explanation:
A Given,
\(\mathrm{y}^2=8 \mathrm{x}=4\). (2) \(\mathrm{x}\)
Double ordinate \(=16\)
Let end points of double ordinate \(=\left(2 t^2 \pm 4 t\right)\)
Length \(=16=8 \mathrm{t}\)
\(\Rightarrow \mathrm{t}=2\)
Endpoint- \(\mathrm{P}(8,-8)\) \& \(\mathrm{Q}(8,8)\)
\(\mathrm{PQ}=\sqrt{(8-8)+(-8-8)^2}=16\)
\(\mathrm{OP}=8 \sqrt{2}\)
\(\mathrm{OQ}=8 \sqrt{2} \quad(\therefore \mathrm{O}\) is origin \()\)
\((\mathrm{OP})^2+(\mathrm{OQ})^2=(\mathrm{PQ})^2\)
\(\therefore\) The angle subtended by double ordinate is ' \(\frac{\pi}{2}\),
AP EAMCET-24.04.2018
Parabola
120093
If a normal chord at a point \(t\) on the parabola \(y^2=4 a x\) subtends a right angle at the vertex then \(t\) equals to
1 1
2 \(\sqrt{2}\)
3 2
4 \(\sqrt{3}\)
Explanation:
B Given,
\(y^2=4 \mathrm{ax}\)
Let, end points \(\left(a t_1^2, 2 a_1\right)\) and \(\left(a t_2^2, 2 a t_2\right)\)
\(\because\) it is normal to parabola :
\(\mathrm{t}_2=-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}\)
Equation of normal to parabola -
\(y=t x+2 a t+a t^3\)
Slope is- \(t_1\)
Now, chord is perpendicular at origin so-
\(\frac{2 \mathrm{at}_1-0}{\mathrm{at}_1^2-0}=\frac{2 \mathrm{at}_1-0}{\mathrm{at}_2^2-0}=-1\)
\(\mathrm{t}_1 \mathrm{t}_2=-4\)
\(\mathrm{t}_1\left(-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}\right)=-4\)
\(\mathrm{t}_1=\sqrt{2}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Parabola
120090
If the straight line \(y=m x+c\) is parallel to the axis of the parabola \(y^2=l x\) and intersects the parabola at \(\left(\frac{c^2}{8}, c\right)\), then the length of the latus rectum is
1 2
2 3
3 4
4 8
Explanation:
D Given, \(\mathrm{y}^2=l \mathrm{x},\left(\frac{\mathrm{c}^2}{8}, \mathrm{c}\right)\)
Length of latus rectum of parabola is \(4 a=l\) We can write -
\(\mathrm{c}^2=l \times \frac{\mathrm{c}^2}{8}\)
\(l=8\)
Also,
\(l=4 \mathrm{a}=8\)
AP EAMCET-2011
Parabola
120091
Let \(M\) be the foot of the perpendicular from a point \(P\) on the parabola \(y^2=8(x-3)\) onto its directrix and let \(S\) be the focus on the parabola. If \(\triangle \mathrm{SPM}\) is an equilateral triangle, then \(P\) is equal to
120092
If the double ordinate of the parabola \(y^2=8 x\) is of Length 16, then the angle subtended by it at the vertex of the parabola is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{3 \pi}{4}\)
4 \(\frac{\pi}{4}\)
Explanation:
A Given,
\(\mathrm{y}^2=8 \mathrm{x}=4\). (2) \(\mathrm{x}\)
Double ordinate \(=16\)
Let end points of double ordinate \(=\left(2 t^2 \pm 4 t\right)\)
Length \(=16=8 \mathrm{t}\)
\(\Rightarrow \mathrm{t}=2\)
Endpoint- \(\mathrm{P}(8,-8)\) \& \(\mathrm{Q}(8,8)\)
\(\mathrm{PQ}=\sqrt{(8-8)+(-8-8)^2}=16\)
\(\mathrm{OP}=8 \sqrt{2}\)
\(\mathrm{OQ}=8 \sqrt{2} \quad(\therefore \mathrm{O}\) is origin \()\)
\((\mathrm{OP})^2+(\mathrm{OQ})^2=(\mathrm{PQ})^2\)
\(\therefore\) The angle subtended by double ordinate is ' \(\frac{\pi}{2}\),
AP EAMCET-24.04.2018
Parabola
120093
If a normal chord at a point \(t\) on the parabola \(y^2=4 a x\) subtends a right angle at the vertex then \(t\) equals to
1 1
2 \(\sqrt{2}\)
3 2
4 \(\sqrt{3}\)
Explanation:
B Given,
\(y^2=4 \mathrm{ax}\)
Let, end points \(\left(a t_1^2, 2 a_1\right)\) and \(\left(a t_2^2, 2 a t_2\right)\)
\(\because\) it is normal to parabola :
\(\mathrm{t}_2=-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}\)
Equation of normal to parabola -
\(y=t x+2 a t+a t^3\)
Slope is- \(t_1\)
Now, chord is perpendicular at origin so-
\(\frac{2 \mathrm{at}_1-0}{\mathrm{at}_1^2-0}=\frac{2 \mathrm{at}_1-0}{\mathrm{at}_2^2-0}=-1\)
\(\mathrm{t}_1 \mathrm{t}_2=-4\)
\(\mathrm{t}_1\left(-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}\right)=-4\)
\(\mathrm{t}_1=\sqrt{2}\)
120090
If the straight line \(y=m x+c\) is parallel to the axis of the parabola \(y^2=l x\) and intersects the parabola at \(\left(\frac{c^2}{8}, c\right)\), then the length of the latus rectum is
1 2
2 3
3 4
4 8
Explanation:
D Given, \(\mathrm{y}^2=l \mathrm{x},\left(\frac{\mathrm{c}^2}{8}, \mathrm{c}\right)\)
Length of latus rectum of parabola is \(4 a=l\) We can write -
\(\mathrm{c}^2=l \times \frac{\mathrm{c}^2}{8}\)
\(l=8\)
Also,
\(l=4 \mathrm{a}=8\)
AP EAMCET-2011
Parabola
120091
Let \(M\) be the foot of the perpendicular from a point \(P\) on the parabola \(y^2=8(x-3)\) onto its directrix and let \(S\) be the focus on the parabola. If \(\triangle \mathrm{SPM}\) is an equilateral triangle, then \(P\) is equal to
120092
If the double ordinate of the parabola \(y^2=8 x\) is of Length 16, then the angle subtended by it at the vertex of the parabola is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{3 \pi}{4}\)
4 \(\frac{\pi}{4}\)
Explanation:
A Given,
\(\mathrm{y}^2=8 \mathrm{x}=4\). (2) \(\mathrm{x}\)
Double ordinate \(=16\)
Let end points of double ordinate \(=\left(2 t^2 \pm 4 t\right)\)
Length \(=16=8 \mathrm{t}\)
\(\Rightarrow \mathrm{t}=2\)
Endpoint- \(\mathrm{P}(8,-8)\) \& \(\mathrm{Q}(8,8)\)
\(\mathrm{PQ}=\sqrt{(8-8)+(-8-8)^2}=16\)
\(\mathrm{OP}=8 \sqrt{2}\)
\(\mathrm{OQ}=8 \sqrt{2} \quad(\therefore \mathrm{O}\) is origin \()\)
\((\mathrm{OP})^2+(\mathrm{OQ})^2=(\mathrm{PQ})^2\)
\(\therefore\) The angle subtended by double ordinate is ' \(\frac{\pi}{2}\),
AP EAMCET-24.04.2018
Parabola
120093
If a normal chord at a point \(t\) on the parabola \(y^2=4 a x\) subtends a right angle at the vertex then \(t\) equals to
1 1
2 \(\sqrt{2}\)
3 2
4 \(\sqrt{3}\)
Explanation:
B Given,
\(y^2=4 \mathrm{ax}\)
Let, end points \(\left(a t_1^2, 2 a_1\right)\) and \(\left(a t_2^2, 2 a t_2\right)\)
\(\because\) it is normal to parabola :
\(\mathrm{t}_2=-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}\)
Equation of normal to parabola -
\(y=t x+2 a t+a t^3\)
Slope is- \(t_1\)
Now, chord is perpendicular at origin so-
\(\frac{2 \mathrm{at}_1-0}{\mathrm{at}_1^2-0}=\frac{2 \mathrm{at}_1-0}{\mathrm{at}_2^2-0}=-1\)
\(\mathrm{t}_1 \mathrm{t}_2=-4\)
\(\mathrm{t}_1\left(-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}\right)=-4\)
\(\mathrm{t}_1=\sqrt{2}\)
