120294
The area (in sq. units) bounded by the parabola \(y=x^2+3\), the tangent to the parabola at \((3,12)\) and the coordinate axes and lying in the first quadrant is
1 6
2 30
3 18
4 24
Explanation:
A Given parabola, \(\mathrm{y}=\mathrm{x}^2+3\)
Tangent of parabola at \((3,12)\) is
\(\frac{\mathrm{Y}+12}{2}=3 \mathrm{x}+3 \Rightarrow \mathrm{Y}=6 \mathrm{x}-6\)
\(\text { Required area } =\int_0^3\left(x^2+3\right) d x-\int_1^3(6 x-6) d x\)
\(=\left[\frac{x^3}{3}+3 x\right]_0^3-\left[3 x^2-6 x\right]_1^3\)
\(=(9+9)-((27-18)+3)=18-12=6\)
TS EAMCET 14.09.2020
Parabola
120295
If \(P(-3,2)\) is an end point of the focal chord \(P Q\) of the parabola \(y^2+4 x+4 y=0\), then the slope of the normal drawn at \(Q\) is
1 \(\frac{-1}{2}\)
2 2
3 \(\frac{1}{2}\)
4 -2
Explanation:
A \(\mathrm{P}(-3,2)\) is an end point of focal chord
\(P Q\) of the parabola \(y^2+4 x+4 y=0\)
\(y^2+4 y+4+4 x-4=0\)
\((y+2)^2=-4(x-1)\)
Let, \(\mathrm{x}-1=-\mathrm{t}^2, \mathrm{y}+2=-2 \mathrm{t} \Rightarrow \mathrm{y}=-2 \mathrm{t}-2\)
\(\Rightarrow \mathrm{x}=1-\mathrm{t}^2, \mathrm{y}=-2 \mathrm{t}-2\)
\(\therefore-3=1-\mathrm{t}^2, 2=-2 \mathrm{t}-2 \Rightarrow \mathrm{t}= \pm 2, \mathrm{t}=-2\)
\(\therefore \mathrm{t}=-2\)
End point of focal chord
\(\therefore \text { point } Q\left(\frac{3}{4},-3\right)\)
\(y^2+4 x+4 y=0 \Rightarrow 2 y \frac{d y}{d x}+4+4 \frac{d y}{d x}=0\)
\(\frac{d y}{d x}=\frac{-4}{2 y+4}=\left(\frac{d y}{d x}\right)_{\left(\frac{3}{4},-3\right)}=\frac{-4}{-2}=2\)
\(\therefore\) Slope of normal of \(\theta=\frac{-1}{2}\)
TS EAMCET-14.09.2020
Parabola
120296
The equation of the common tangent of the parabolas \(\mathrm{x}^2=108 \mathrm{y}\) and \(\mathrm{y}^2=32 \mathrm{x}\) is
1 \(2 x+3 y+36=0\)
2 \(2 x+3 y=36\)
3 \(3 x+2 y+36=0\)
4 \(3 x+2 y=36\)
Explanation:
A Given equation of parabola
\(x^2=108 y(4 a=108 \Rightarrow a=27)\)
Equation of tangent, \(\mathrm{y}=\mathrm{mx}-27 \mathrm{~m}^2\)
And equation of parabola
\(y^2=32 x(4 a=32 \Rightarrow a=8)\)
Equation of tangent, \(y=m x+\frac{8}{m}\)
Since, equation (i) and (ii) are identical,
\(m x-27 m^2=m x+\frac{8}{m}\)
\(-27 m^2=\frac{8}{m} \Rightarrow m^3=\frac{-8}{27} \Rightarrow m=\frac{-2}{3}\)
\(\therefore\) Required equation of tangent
\(y=\left(\frac{-2}{3}\right) x+\frac{8}{\frac{-2}{3}}\)
\(y=\frac{-2 x}{3}-12\)
\(3 y=-2 x-36\)
\(2 x+3 y+36=0\)
TS EAMCET-10.09.2020
Parabola
120297
If the normal at one end of the latusrectum of the parabola \(y^2=16 x\) meets the \(X\)-axis at the point \(P\), then the length of the chord passing through \(P\) and perpendicular to the normal is.
1 \(48 \sqrt{2}\)
2 \(32 \sqrt{2}\)
3 \(24 \sqrt{2}\)
4 \(20 \sqrt{2}\)
Explanation:
B Given,
\(\mathbf{y}^2=16 \mathrm{x}\)
Coordinate of latusrectum \((4,8)\)
Equation of normal at \((4,8)\) is \(y=-x+12\)
It cuts \(\mathrm{x}-\) axis at \((12,0)\)
|Equation of chord passing through \((12,0)\) and perpendicular of normal is
\(\mathrm{y}=\mathrm{x}-12\)
Put the value of \(y\) in \(y^2=16 x\), we get-
\((12-x)^2=16 x\)
\(144-24 \mathrm{x}+\mathrm{x}^2=16 \mathrm{x}\)
\(\mathrm{x}^2-40 \mathrm{x}+144=0\)
\((\mathrm{x}-36)(\mathrm{x}-4)=0\)
\(\mathrm{x}=4,36\)
\(\therefore \quad \mathrm{y}=4-12=-8 \text { and } \mathrm{y}=36-12=24\)
\(\therefore \text { Length of chord }=\sqrt{(36-4)^2+(24+8)^2}\)
\(=\sqrt{32^2+32^2}=32 \sqrt{2}\)
TS EAMCET-04.05.2019
Parabola
120298
If the line \(x-y=-4 K\) is a tangent to the parabola \(y^2=8 \mathrm{x}\) at \(P\), then the perpendicular distance of normal at \(P\) from \((\mathrm{K}, 2 \mathrm{~K})\) is
1 \(\frac{5}{2 \sqrt{2}}\)
2 \(\frac{7}{2 \sqrt{2}}\)
3 \(\frac{9}{2 \sqrt{2}}\)
4 \(\frac{1}{2 \sqrt{2}}\)
Explanation:
C Given,
\(\mathrm{x}-\mathrm{y}=-4 \mathrm{k}\)
\(y=x+4 k\) is a tangent to the parabola \(y^2=8 x\),
Therefore, \(4 \mathrm{k}=\frac{2}{1}\)
[Here \(4 \mathrm{a}=8, \mathrm{a}=2\) ]
\(\mathrm{k}=\frac{1}{2}\)
Also point of contact \(\mathrm{p}\) is \(\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right)=(2,4)\)
Now, equation of normal to the \(y^2=8 x\) at \((2,4)\) is
\((y-4)=\frac{-4}{2(2)}(x-2)\)
\(\left[\because\right.\) Equation of normal to the parabola \(y^2=4\) are at \(\left(x_1\right.\), \(\left.\mathrm{y}_1\right)\) is \(\left.\mathrm{y}-\mathrm{y}_1=\frac{-\mathrm{b}}{2 \mathrm{a}}\left(\mathrm{x}-\mathrm{x}_1\right)\right]\)
\(y-4=-1(x-2)\)
\(y-4=-x+2\)
\(x+y=6\)
\(x+y-6=0\)
The perpendicular distance of normal from \((\mathrm{k}, 2 \mathrm{k})\) i.e. \(\left(\frac{1}{2}, 1\right)\)
\(=\frac{\left|\frac{1}{2}+1-6\right|}{\sqrt{1^2+1^2}}=\frac{\left|\frac{3}{2}-6\right|}{\sqrt{2}}=\frac{9}{2 \sqrt{2}}\)
120294
The area (in sq. units) bounded by the parabola \(y=x^2+3\), the tangent to the parabola at \((3,12)\) and the coordinate axes and lying in the first quadrant is
1 6
2 30
3 18
4 24
Explanation:
A Given parabola, \(\mathrm{y}=\mathrm{x}^2+3\)
Tangent of parabola at \((3,12)\) is
\(\frac{\mathrm{Y}+12}{2}=3 \mathrm{x}+3 \Rightarrow \mathrm{Y}=6 \mathrm{x}-6\)
\(\text { Required area } =\int_0^3\left(x^2+3\right) d x-\int_1^3(6 x-6) d x\)
\(=\left[\frac{x^3}{3}+3 x\right]_0^3-\left[3 x^2-6 x\right]_1^3\)
\(=(9+9)-((27-18)+3)=18-12=6\)
TS EAMCET 14.09.2020
Parabola
120295
If \(P(-3,2)\) is an end point of the focal chord \(P Q\) of the parabola \(y^2+4 x+4 y=0\), then the slope of the normal drawn at \(Q\) is
1 \(\frac{-1}{2}\)
2 2
3 \(\frac{1}{2}\)
4 -2
Explanation:
A \(\mathrm{P}(-3,2)\) is an end point of focal chord
\(P Q\) of the parabola \(y^2+4 x+4 y=0\)
\(y^2+4 y+4+4 x-4=0\)
\((y+2)^2=-4(x-1)\)
Let, \(\mathrm{x}-1=-\mathrm{t}^2, \mathrm{y}+2=-2 \mathrm{t} \Rightarrow \mathrm{y}=-2 \mathrm{t}-2\)
\(\Rightarrow \mathrm{x}=1-\mathrm{t}^2, \mathrm{y}=-2 \mathrm{t}-2\)
\(\therefore-3=1-\mathrm{t}^2, 2=-2 \mathrm{t}-2 \Rightarrow \mathrm{t}= \pm 2, \mathrm{t}=-2\)
\(\therefore \mathrm{t}=-2\)
End point of focal chord
\(\therefore \text { point } Q\left(\frac{3}{4},-3\right)\)
\(y^2+4 x+4 y=0 \Rightarrow 2 y \frac{d y}{d x}+4+4 \frac{d y}{d x}=0\)
\(\frac{d y}{d x}=\frac{-4}{2 y+4}=\left(\frac{d y}{d x}\right)_{\left(\frac{3}{4},-3\right)}=\frac{-4}{-2}=2\)
\(\therefore\) Slope of normal of \(\theta=\frac{-1}{2}\)
TS EAMCET-14.09.2020
Parabola
120296
The equation of the common tangent of the parabolas \(\mathrm{x}^2=108 \mathrm{y}\) and \(\mathrm{y}^2=32 \mathrm{x}\) is
1 \(2 x+3 y+36=0\)
2 \(2 x+3 y=36\)
3 \(3 x+2 y+36=0\)
4 \(3 x+2 y=36\)
Explanation:
A Given equation of parabola
\(x^2=108 y(4 a=108 \Rightarrow a=27)\)
Equation of tangent, \(\mathrm{y}=\mathrm{mx}-27 \mathrm{~m}^2\)
And equation of parabola
\(y^2=32 x(4 a=32 \Rightarrow a=8)\)
Equation of tangent, \(y=m x+\frac{8}{m}\)
Since, equation (i) and (ii) are identical,
\(m x-27 m^2=m x+\frac{8}{m}\)
\(-27 m^2=\frac{8}{m} \Rightarrow m^3=\frac{-8}{27} \Rightarrow m=\frac{-2}{3}\)
\(\therefore\) Required equation of tangent
\(y=\left(\frac{-2}{3}\right) x+\frac{8}{\frac{-2}{3}}\)
\(y=\frac{-2 x}{3}-12\)
\(3 y=-2 x-36\)
\(2 x+3 y+36=0\)
TS EAMCET-10.09.2020
Parabola
120297
If the normal at one end of the latusrectum of the parabola \(y^2=16 x\) meets the \(X\)-axis at the point \(P\), then the length of the chord passing through \(P\) and perpendicular to the normal is.
1 \(48 \sqrt{2}\)
2 \(32 \sqrt{2}\)
3 \(24 \sqrt{2}\)
4 \(20 \sqrt{2}\)
Explanation:
B Given,
\(\mathbf{y}^2=16 \mathrm{x}\)
Coordinate of latusrectum \((4,8)\)
Equation of normal at \((4,8)\) is \(y=-x+12\)
It cuts \(\mathrm{x}-\) axis at \((12,0)\)
|Equation of chord passing through \((12,0)\) and perpendicular of normal is
\(\mathrm{y}=\mathrm{x}-12\)
Put the value of \(y\) in \(y^2=16 x\), we get-
\((12-x)^2=16 x\)
\(144-24 \mathrm{x}+\mathrm{x}^2=16 \mathrm{x}\)
\(\mathrm{x}^2-40 \mathrm{x}+144=0\)
\((\mathrm{x}-36)(\mathrm{x}-4)=0\)
\(\mathrm{x}=4,36\)
\(\therefore \quad \mathrm{y}=4-12=-8 \text { and } \mathrm{y}=36-12=24\)
\(\therefore \text { Length of chord }=\sqrt{(36-4)^2+(24+8)^2}\)
\(=\sqrt{32^2+32^2}=32 \sqrt{2}\)
TS EAMCET-04.05.2019
Parabola
120298
If the line \(x-y=-4 K\) is a tangent to the parabola \(y^2=8 \mathrm{x}\) at \(P\), then the perpendicular distance of normal at \(P\) from \((\mathrm{K}, 2 \mathrm{~K})\) is
1 \(\frac{5}{2 \sqrt{2}}\)
2 \(\frac{7}{2 \sqrt{2}}\)
3 \(\frac{9}{2 \sqrt{2}}\)
4 \(\frac{1}{2 \sqrt{2}}\)
Explanation:
C Given,
\(\mathrm{x}-\mathrm{y}=-4 \mathrm{k}\)
\(y=x+4 k\) is a tangent to the parabola \(y^2=8 x\),
Therefore, \(4 \mathrm{k}=\frac{2}{1}\)
[Here \(4 \mathrm{a}=8, \mathrm{a}=2\) ]
\(\mathrm{k}=\frac{1}{2}\)
Also point of contact \(\mathrm{p}\) is \(\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right)=(2,4)\)
Now, equation of normal to the \(y^2=8 x\) at \((2,4)\) is
\((y-4)=\frac{-4}{2(2)}(x-2)\)
\(\left[\because\right.\) Equation of normal to the parabola \(y^2=4\) are at \(\left(x_1\right.\), \(\left.\mathrm{y}_1\right)\) is \(\left.\mathrm{y}-\mathrm{y}_1=\frac{-\mathrm{b}}{2 \mathrm{a}}\left(\mathrm{x}-\mathrm{x}_1\right)\right]\)
\(y-4=-1(x-2)\)
\(y-4=-x+2\)
\(x+y=6\)
\(x+y-6=0\)
The perpendicular distance of normal from \((\mathrm{k}, 2 \mathrm{k})\) i.e. \(\left(\frac{1}{2}, 1\right)\)
\(=\frac{\left|\frac{1}{2}+1-6\right|}{\sqrt{1^2+1^2}}=\frac{\left|\frac{3}{2}-6\right|}{\sqrt{2}}=\frac{9}{2 \sqrt{2}}\)
120294
The area (in sq. units) bounded by the parabola \(y=x^2+3\), the tangent to the parabola at \((3,12)\) and the coordinate axes and lying in the first quadrant is
1 6
2 30
3 18
4 24
Explanation:
A Given parabola, \(\mathrm{y}=\mathrm{x}^2+3\)
Tangent of parabola at \((3,12)\) is
\(\frac{\mathrm{Y}+12}{2}=3 \mathrm{x}+3 \Rightarrow \mathrm{Y}=6 \mathrm{x}-6\)
\(\text { Required area } =\int_0^3\left(x^2+3\right) d x-\int_1^3(6 x-6) d x\)
\(=\left[\frac{x^3}{3}+3 x\right]_0^3-\left[3 x^2-6 x\right]_1^3\)
\(=(9+9)-((27-18)+3)=18-12=6\)
TS EAMCET 14.09.2020
Parabola
120295
If \(P(-3,2)\) is an end point of the focal chord \(P Q\) of the parabola \(y^2+4 x+4 y=0\), then the slope of the normal drawn at \(Q\) is
1 \(\frac{-1}{2}\)
2 2
3 \(\frac{1}{2}\)
4 -2
Explanation:
A \(\mathrm{P}(-3,2)\) is an end point of focal chord
\(P Q\) of the parabola \(y^2+4 x+4 y=0\)
\(y^2+4 y+4+4 x-4=0\)
\((y+2)^2=-4(x-1)\)
Let, \(\mathrm{x}-1=-\mathrm{t}^2, \mathrm{y}+2=-2 \mathrm{t} \Rightarrow \mathrm{y}=-2 \mathrm{t}-2\)
\(\Rightarrow \mathrm{x}=1-\mathrm{t}^2, \mathrm{y}=-2 \mathrm{t}-2\)
\(\therefore-3=1-\mathrm{t}^2, 2=-2 \mathrm{t}-2 \Rightarrow \mathrm{t}= \pm 2, \mathrm{t}=-2\)
\(\therefore \mathrm{t}=-2\)
End point of focal chord
\(\therefore \text { point } Q\left(\frac{3}{4},-3\right)\)
\(y^2+4 x+4 y=0 \Rightarrow 2 y \frac{d y}{d x}+4+4 \frac{d y}{d x}=0\)
\(\frac{d y}{d x}=\frac{-4}{2 y+4}=\left(\frac{d y}{d x}\right)_{\left(\frac{3}{4},-3\right)}=\frac{-4}{-2}=2\)
\(\therefore\) Slope of normal of \(\theta=\frac{-1}{2}\)
TS EAMCET-14.09.2020
Parabola
120296
The equation of the common tangent of the parabolas \(\mathrm{x}^2=108 \mathrm{y}\) and \(\mathrm{y}^2=32 \mathrm{x}\) is
1 \(2 x+3 y+36=0\)
2 \(2 x+3 y=36\)
3 \(3 x+2 y+36=0\)
4 \(3 x+2 y=36\)
Explanation:
A Given equation of parabola
\(x^2=108 y(4 a=108 \Rightarrow a=27)\)
Equation of tangent, \(\mathrm{y}=\mathrm{mx}-27 \mathrm{~m}^2\)
And equation of parabola
\(y^2=32 x(4 a=32 \Rightarrow a=8)\)
Equation of tangent, \(y=m x+\frac{8}{m}\)
Since, equation (i) and (ii) are identical,
\(m x-27 m^2=m x+\frac{8}{m}\)
\(-27 m^2=\frac{8}{m} \Rightarrow m^3=\frac{-8}{27} \Rightarrow m=\frac{-2}{3}\)
\(\therefore\) Required equation of tangent
\(y=\left(\frac{-2}{3}\right) x+\frac{8}{\frac{-2}{3}}\)
\(y=\frac{-2 x}{3}-12\)
\(3 y=-2 x-36\)
\(2 x+3 y+36=0\)
TS EAMCET-10.09.2020
Parabola
120297
If the normal at one end of the latusrectum of the parabola \(y^2=16 x\) meets the \(X\)-axis at the point \(P\), then the length of the chord passing through \(P\) and perpendicular to the normal is.
1 \(48 \sqrt{2}\)
2 \(32 \sqrt{2}\)
3 \(24 \sqrt{2}\)
4 \(20 \sqrt{2}\)
Explanation:
B Given,
\(\mathbf{y}^2=16 \mathrm{x}\)
Coordinate of latusrectum \((4,8)\)
Equation of normal at \((4,8)\) is \(y=-x+12\)
It cuts \(\mathrm{x}-\) axis at \((12,0)\)
|Equation of chord passing through \((12,0)\) and perpendicular of normal is
\(\mathrm{y}=\mathrm{x}-12\)
Put the value of \(y\) in \(y^2=16 x\), we get-
\((12-x)^2=16 x\)
\(144-24 \mathrm{x}+\mathrm{x}^2=16 \mathrm{x}\)
\(\mathrm{x}^2-40 \mathrm{x}+144=0\)
\((\mathrm{x}-36)(\mathrm{x}-4)=0\)
\(\mathrm{x}=4,36\)
\(\therefore \quad \mathrm{y}=4-12=-8 \text { and } \mathrm{y}=36-12=24\)
\(\therefore \text { Length of chord }=\sqrt{(36-4)^2+(24+8)^2}\)
\(=\sqrt{32^2+32^2}=32 \sqrt{2}\)
TS EAMCET-04.05.2019
Parabola
120298
If the line \(x-y=-4 K\) is a tangent to the parabola \(y^2=8 \mathrm{x}\) at \(P\), then the perpendicular distance of normal at \(P\) from \((\mathrm{K}, 2 \mathrm{~K})\) is
1 \(\frac{5}{2 \sqrt{2}}\)
2 \(\frac{7}{2 \sqrt{2}}\)
3 \(\frac{9}{2 \sqrt{2}}\)
4 \(\frac{1}{2 \sqrt{2}}\)
Explanation:
C Given,
\(\mathrm{x}-\mathrm{y}=-4 \mathrm{k}\)
\(y=x+4 k\) is a tangent to the parabola \(y^2=8 x\),
Therefore, \(4 \mathrm{k}=\frac{2}{1}\)
[Here \(4 \mathrm{a}=8, \mathrm{a}=2\) ]
\(\mathrm{k}=\frac{1}{2}\)
Also point of contact \(\mathrm{p}\) is \(\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right)=(2,4)\)
Now, equation of normal to the \(y^2=8 x\) at \((2,4)\) is
\((y-4)=\frac{-4}{2(2)}(x-2)\)
\(\left[\because\right.\) Equation of normal to the parabola \(y^2=4\) are at \(\left(x_1\right.\), \(\left.\mathrm{y}_1\right)\) is \(\left.\mathrm{y}-\mathrm{y}_1=\frac{-\mathrm{b}}{2 \mathrm{a}}\left(\mathrm{x}-\mathrm{x}_1\right)\right]\)
\(y-4=-1(x-2)\)
\(y-4=-x+2\)
\(x+y=6\)
\(x+y-6=0\)
The perpendicular distance of normal from \((\mathrm{k}, 2 \mathrm{k})\) i.e. \(\left(\frac{1}{2}, 1\right)\)
\(=\frac{\left|\frac{1}{2}+1-6\right|}{\sqrt{1^2+1^2}}=\frac{\left|\frac{3}{2}-6\right|}{\sqrt{2}}=\frac{9}{2 \sqrt{2}}\)
120294
The area (in sq. units) bounded by the parabola \(y=x^2+3\), the tangent to the parabola at \((3,12)\) and the coordinate axes and lying in the first quadrant is
1 6
2 30
3 18
4 24
Explanation:
A Given parabola, \(\mathrm{y}=\mathrm{x}^2+3\)
Tangent of parabola at \((3,12)\) is
\(\frac{\mathrm{Y}+12}{2}=3 \mathrm{x}+3 \Rightarrow \mathrm{Y}=6 \mathrm{x}-6\)
\(\text { Required area } =\int_0^3\left(x^2+3\right) d x-\int_1^3(6 x-6) d x\)
\(=\left[\frac{x^3}{3}+3 x\right]_0^3-\left[3 x^2-6 x\right]_1^3\)
\(=(9+9)-((27-18)+3)=18-12=6\)
TS EAMCET 14.09.2020
Parabola
120295
If \(P(-3,2)\) is an end point of the focal chord \(P Q\) of the parabola \(y^2+4 x+4 y=0\), then the slope of the normal drawn at \(Q\) is
1 \(\frac{-1}{2}\)
2 2
3 \(\frac{1}{2}\)
4 -2
Explanation:
A \(\mathrm{P}(-3,2)\) is an end point of focal chord
\(P Q\) of the parabola \(y^2+4 x+4 y=0\)
\(y^2+4 y+4+4 x-4=0\)
\((y+2)^2=-4(x-1)\)
Let, \(\mathrm{x}-1=-\mathrm{t}^2, \mathrm{y}+2=-2 \mathrm{t} \Rightarrow \mathrm{y}=-2 \mathrm{t}-2\)
\(\Rightarrow \mathrm{x}=1-\mathrm{t}^2, \mathrm{y}=-2 \mathrm{t}-2\)
\(\therefore-3=1-\mathrm{t}^2, 2=-2 \mathrm{t}-2 \Rightarrow \mathrm{t}= \pm 2, \mathrm{t}=-2\)
\(\therefore \mathrm{t}=-2\)
End point of focal chord
\(\therefore \text { point } Q\left(\frac{3}{4},-3\right)\)
\(y^2+4 x+4 y=0 \Rightarrow 2 y \frac{d y}{d x}+4+4 \frac{d y}{d x}=0\)
\(\frac{d y}{d x}=\frac{-4}{2 y+4}=\left(\frac{d y}{d x}\right)_{\left(\frac{3}{4},-3\right)}=\frac{-4}{-2}=2\)
\(\therefore\) Slope of normal of \(\theta=\frac{-1}{2}\)
TS EAMCET-14.09.2020
Parabola
120296
The equation of the common tangent of the parabolas \(\mathrm{x}^2=108 \mathrm{y}\) and \(\mathrm{y}^2=32 \mathrm{x}\) is
1 \(2 x+3 y+36=0\)
2 \(2 x+3 y=36\)
3 \(3 x+2 y+36=0\)
4 \(3 x+2 y=36\)
Explanation:
A Given equation of parabola
\(x^2=108 y(4 a=108 \Rightarrow a=27)\)
Equation of tangent, \(\mathrm{y}=\mathrm{mx}-27 \mathrm{~m}^2\)
And equation of parabola
\(y^2=32 x(4 a=32 \Rightarrow a=8)\)
Equation of tangent, \(y=m x+\frac{8}{m}\)
Since, equation (i) and (ii) are identical,
\(m x-27 m^2=m x+\frac{8}{m}\)
\(-27 m^2=\frac{8}{m} \Rightarrow m^3=\frac{-8}{27} \Rightarrow m=\frac{-2}{3}\)
\(\therefore\) Required equation of tangent
\(y=\left(\frac{-2}{3}\right) x+\frac{8}{\frac{-2}{3}}\)
\(y=\frac{-2 x}{3}-12\)
\(3 y=-2 x-36\)
\(2 x+3 y+36=0\)
TS EAMCET-10.09.2020
Parabola
120297
If the normal at one end of the latusrectum of the parabola \(y^2=16 x\) meets the \(X\)-axis at the point \(P\), then the length of the chord passing through \(P\) and perpendicular to the normal is.
1 \(48 \sqrt{2}\)
2 \(32 \sqrt{2}\)
3 \(24 \sqrt{2}\)
4 \(20 \sqrt{2}\)
Explanation:
B Given,
\(\mathbf{y}^2=16 \mathrm{x}\)
Coordinate of latusrectum \((4,8)\)
Equation of normal at \((4,8)\) is \(y=-x+12\)
It cuts \(\mathrm{x}-\) axis at \((12,0)\)
|Equation of chord passing through \((12,0)\) and perpendicular of normal is
\(\mathrm{y}=\mathrm{x}-12\)
Put the value of \(y\) in \(y^2=16 x\), we get-
\((12-x)^2=16 x\)
\(144-24 \mathrm{x}+\mathrm{x}^2=16 \mathrm{x}\)
\(\mathrm{x}^2-40 \mathrm{x}+144=0\)
\((\mathrm{x}-36)(\mathrm{x}-4)=0\)
\(\mathrm{x}=4,36\)
\(\therefore \quad \mathrm{y}=4-12=-8 \text { and } \mathrm{y}=36-12=24\)
\(\therefore \text { Length of chord }=\sqrt{(36-4)^2+(24+8)^2}\)
\(=\sqrt{32^2+32^2}=32 \sqrt{2}\)
TS EAMCET-04.05.2019
Parabola
120298
If the line \(x-y=-4 K\) is a tangent to the parabola \(y^2=8 \mathrm{x}\) at \(P\), then the perpendicular distance of normal at \(P\) from \((\mathrm{K}, 2 \mathrm{~K})\) is
1 \(\frac{5}{2 \sqrt{2}}\)
2 \(\frac{7}{2 \sqrt{2}}\)
3 \(\frac{9}{2 \sqrt{2}}\)
4 \(\frac{1}{2 \sqrt{2}}\)
Explanation:
C Given,
\(\mathrm{x}-\mathrm{y}=-4 \mathrm{k}\)
\(y=x+4 k\) is a tangent to the parabola \(y^2=8 x\),
Therefore, \(4 \mathrm{k}=\frac{2}{1}\)
[Here \(4 \mathrm{a}=8, \mathrm{a}=2\) ]
\(\mathrm{k}=\frac{1}{2}\)
Also point of contact \(\mathrm{p}\) is \(\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right)=(2,4)\)
Now, equation of normal to the \(y^2=8 x\) at \((2,4)\) is
\((y-4)=\frac{-4}{2(2)}(x-2)\)
\(\left[\because\right.\) Equation of normal to the parabola \(y^2=4\) are at \(\left(x_1\right.\), \(\left.\mathrm{y}_1\right)\) is \(\left.\mathrm{y}-\mathrm{y}_1=\frac{-\mathrm{b}}{2 \mathrm{a}}\left(\mathrm{x}-\mathrm{x}_1\right)\right]\)
\(y-4=-1(x-2)\)
\(y-4=-x+2\)
\(x+y=6\)
\(x+y-6=0\)
The perpendicular distance of normal from \((\mathrm{k}, 2 \mathrm{k})\) i.e. \(\left(\frac{1}{2}, 1\right)\)
\(=\frac{\left|\frac{1}{2}+1-6\right|}{\sqrt{1^2+1^2}}=\frac{\left|\frac{3}{2}-6\right|}{\sqrt{2}}=\frac{9}{2 \sqrt{2}}\)
120294
The area (in sq. units) bounded by the parabola \(y=x^2+3\), the tangent to the parabola at \((3,12)\) and the coordinate axes and lying in the first quadrant is
1 6
2 30
3 18
4 24
Explanation:
A Given parabola, \(\mathrm{y}=\mathrm{x}^2+3\)
Tangent of parabola at \((3,12)\) is
\(\frac{\mathrm{Y}+12}{2}=3 \mathrm{x}+3 \Rightarrow \mathrm{Y}=6 \mathrm{x}-6\)
\(\text { Required area } =\int_0^3\left(x^2+3\right) d x-\int_1^3(6 x-6) d x\)
\(=\left[\frac{x^3}{3}+3 x\right]_0^3-\left[3 x^2-6 x\right]_1^3\)
\(=(9+9)-((27-18)+3)=18-12=6\)
TS EAMCET 14.09.2020
Parabola
120295
If \(P(-3,2)\) is an end point of the focal chord \(P Q\) of the parabola \(y^2+4 x+4 y=0\), then the slope of the normal drawn at \(Q\) is
1 \(\frac{-1}{2}\)
2 2
3 \(\frac{1}{2}\)
4 -2
Explanation:
A \(\mathrm{P}(-3,2)\) is an end point of focal chord
\(P Q\) of the parabola \(y^2+4 x+4 y=0\)
\(y^2+4 y+4+4 x-4=0\)
\((y+2)^2=-4(x-1)\)
Let, \(\mathrm{x}-1=-\mathrm{t}^2, \mathrm{y}+2=-2 \mathrm{t} \Rightarrow \mathrm{y}=-2 \mathrm{t}-2\)
\(\Rightarrow \mathrm{x}=1-\mathrm{t}^2, \mathrm{y}=-2 \mathrm{t}-2\)
\(\therefore-3=1-\mathrm{t}^2, 2=-2 \mathrm{t}-2 \Rightarrow \mathrm{t}= \pm 2, \mathrm{t}=-2\)
\(\therefore \mathrm{t}=-2\)
End point of focal chord
\(\therefore \text { point } Q\left(\frac{3}{4},-3\right)\)
\(y^2+4 x+4 y=0 \Rightarrow 2 y \frac{d y}{d x}+4+4 \frac{d y}{d x}=0\)
\(\frac{d y}{d x}=\frac{-4}{2 y+4}=\left(\frac{d y}{d x}\right)_{\left(\frac{3}{4},-3\right)}=\frac{-4}{-2}=2\)
\(\therefore\) Slope of normal of \(\theta=\frac{-1}{2}\)
TS EAMCET-14.09.2020
Parabola
120296
The equation of the common tangent of the parabolas \(\mathrm{x}^2=108 \mathrm{y}\) and \(\mathrm{y}^2=32 \mathrm{x}\) is
1 \(2 x+3 y+36=0\)
2 \(2 x+3 y=36\)
3 \(3 x+2 y+36=0\)
4 \(3 x+2 y=36\)
Explanation:
A Given equation of parabola
\(x^2=108 y(4 a=108 \Rightarrow a=27)\)
Equation of tangent, \(\mathrm{y}=\mathrm{mx}-27 \mathrm{~m}^2\)
And equation of parabola
\(y^2=32 x(4 a=32 \Rightarrow a=8)\)
Equation of tangent, \(y=m x+\frac{8}{m}\)
Since, equation (i) and (ii) are identical,
\(m x-27 m^2=m x+\frac{8}{m}\)
\(-27 m^2=\frac{8}{m} \Rightarrow m^3=\frac{-8}{27} \Rightarrow m=\frac{-2}{3}\)
\(\therefore\) Required equation of tangent
\(y=\left(\frac{-2}{3}\right) x+\frac{8}{\frac{-2}{3}}\)
\(y=\frac{-2 x}{3}-12\)
\(3 y=-2 x-36\)
\(2 x+3 y+36=0\)
TS EAMCET-10.09.2020
Parabola
120297
If the normal at one end of the latusrectum of the parabola \(y^2=16 x\) meets the \(X\)-axis at the point \(P\), then the length of the chord passing through \(P\) and perpendicular to the normal is.
1 \(48 \sqrt{2}\)
2 \(32 \sqrt{2}\)
3 \(24 \sqrt{2}\)
4 \(20 \sqrt{2}\)
Explanation:
B Given,
\(\mathbf{y}^2=16 \mathrm{x}\)
Coordinate of latusrectum \((4,8)\)
Equation of normal at \((4,8)\) is \(y=-x+12\)
It cuts \(\mathrm{x}-\) axis at \((12,0)\)
|Equation of chord passing through \((12,0)\) and perpendicular of normal is
\(\mathrm{y}=\mathrm{x}-12\)
Put the value of \(y\) in \(y^2=16 x\), we get-
\((12-x)^2=16 x\)
\(144-24 \mathrm{x}+\mathrm{x}^2=16 \mathrm{x}\)
\(\mathrm{x}^2-40 \mathrm{x}+144=0\)
\((\mathrm{x}-36)(\mathrm{x}-4)=0\)
\(\mathrm{x}=4,36\)
\(\therefore \quad \mathrm{y}=4-12=-8 \text { and } \mathrm{y}=36-12=24\)
\(\therefore \text { Length of chord }=\sqrt{(36-4)^2+(24+8)^2}\)
\(=\sqrt{32^2+32^2}=32 \sqrt{2}\)
TS EAMCET-04.05.2019
Parabola
120298
If the line \(x-y=-4 K\) is a tangent to the parabola \(y^2=8 \mathrm{x}\) at \(P\), then the perpendicular distance of normal at \(P\) from \((\mathrm{K}, 2 \mathrm{~K})\) is
1 \(\frac{5}{2 \sqrt{2}}\)
2 \(\frac{7}{2 \sqrt{2}}\)
3 \(\frac{9}{2 \sqrt{2}}\)
4 \(\frac{1}{2 \sqrt{2}}\)
Explanation:
C Given,
\(\mathrm{x}-\mathrm{y}=-4 \mathrm{k}\)
\(y=x+4 k\) is a tangent to the parabola \(y^2=8 x\),
Therefore, \(4 \mathrm{k}=\frac{2}{1}\)
[Here \(4 \mathrm{a}=8, \mathrm{a}=2\) ]
\(\mathrm{k}=\frac{1}{2}\)
Also point of contact \(\mathrm{p}\) is \(\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right)=(2,4)\)
Now, equation of normal to the \(y^2=8 x\) at \((2,4)\) is
\((y-4)=\frac{-4}{2(2)}(x-2)\)
\(\left[\because\right.\) Equation of normal to the parabola \(y^2=4\) are at \(\left(x_1\right.\), \(\left.\mathrm{y}_1\right)\) is \(\left.\mathrm{y}-\mathrm{y}_1=\frac{-\mathrm{b}}{2 \mathrm{a}}\left(\mathrm{x}-\mathrm{x}_1\right)\right]\)
\(y-4=-1(x-2)\)
\(y-4=-x+2\)
\(x+y=6\)
\(x+y-6=0\)
The perpendicular distance of normal from \((\mathrm{k}, 2 \mathrm{k})\) i.e. \(\left(\frac{1}{2}, 1\right)\)
\(=\frac{\left|\frac{1}{2}+1-6\right|}{\sqrt{1^2+1^2}}=\frac{\left|\frac{3}{2}-6\right|}{\sqrt{2}}=\frac{9}{2 \sqrt{2}}\)
