120244
The equations of two sides of a variable triangle are \(x=0\) and \(y=3\), and its third side is a tangent to the parabola \(y^2=6 x\). The locus of its circum centre is:
1 \(4 y^2-18 y+3 x+18=0\)
2 \(4 y^2-18 y-3 x+18=0\)
3 \(4 y^2+18 y+3 x+18=0\)
4 \(4 y^2-18 y-3 x-18=0\)
Explanation:
A Given,
\(y^2=6 x\)
\(y^2=4 \times\left(\frac{3}{2}\right) x\)
\(a=\frac{3}{2}\)
Now, equation of line :
\(\mathrm{y}=\mathrm{mx}+\frac{3}{2 \mathrm{~m}}\)
\(\mathrm{h}=\frac{6 \mathrm{~m}-3}{4 \mathrm{~m}^2}, \mathrm{k}=\frac{6 \mathrm{~m}+3}{4 \mathrm{~m}}\)
Now, eliminating ' \(\mathrm{m}\) ' we get -
\(3 \mathrm{x}=2\left(-2 \mathrm{k}^2+9 \mathrm{k}-9\right)\)
\(4 \mathrm{y}^2-18 \mathrm{y}+3 \mathrm{x}+18=0\)
JEE Main-25.01.2023
Parabola
120245
The parabolas : \(\mathrm{ax}^2+2 \mathrm{bx}+\mathrm{cy}=0\) and \(\mathrm{dx}^2+\) \(2 e x+f y=0\) intersect on the line \(y=1\). If a, b, \(c\), \(d, e, f\) are positive real numbers and \(a, b, c\), are in G.P., then
1 \(\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in G.P
2 d, e, f are in A.P
3 d, e, f are in G.P
4 \(\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in A.P
Explanation:
D Given,
\(\mathrm{P}_1: \mathrm{ax}^2+2 \mathrm{bx}+\mathrm{cy}=0\)
\(\mathrm{P}_2: d \mathrm{dx}^2+2 \mathrm{ex}+\mathrm{fy}=0\)
From \(\mathrm{P}_1: \mathrm{ax}^2+2 \sqrt{\mathrm{ac}} \mathrm{x}+\mathrm{cy}=0 \quad\left\{\because \mathrm{b}^2=\mathrm{ac}\right\}\)
\((x \sqrt{a}+\sqrt{c})^2=0\)
\(x=-\sqrt{\frac{c}{a}}\)
From, \(\mathrm{P}_2: \mathrm{d}\left(\frac{\mathrm{c}}{\mathrm{a}}\right)+2 \mathrm{e}\left(-\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\right)+\mathrm{f}=0\) from \(\left.=(\mathrm{i})\right\}\)
\(\Rightarrow \frac{\mathrm{dc}}{\mathrm{a}}+\mathrm{f}=2 \mathrm{e} \sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\)
\(\Rightarrow \frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{f}}{\mathrm{c}}=2 \mathrm{e} \sqrt{\frac{1}{\mathrm{ac}}}\)
\(\Rightarrow \frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{f}}{\mathrm{c}}=\frac{2 \mathrm{e}}{\mathrm{b}}\)
\(\therefore \quad \frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in AP.
JEE Main-30.01.2023
Parabola
120246
The tangents at the points \(A(1,3)\) and \(B(1,-1)\) on the parabola \(y^2-2 x-2 y=1\) meet at the point \(P\). Then the area (in unit \({ }^2\) ) of the triangle PAB is :
1 4
2 6
3 7
4 8
Explanation:
D The given parabola is
\(y^2-2 x-2 y=1\)
\((y-1)^2=2(x+1)\)
Or
Now, from the above given information the tangent at the point \(A(1,3)\) on the parabola \(y^2-2 x-2 y=1\) is
\(y(3)-(x+1)-(y+3)=1\)
\(3 y-x-1-y-3=1\)
\(2 y-x=5\)
And tangent at \((1,-1)\) is
\(y(-1)-(x+1)-(y-1)=1\)
\(-y-x-1-y+1=1\)
\(x+2 y=-1\)
Now, solving equitation (i) and (ii), we get-
\(x=-3 \text { and } y=1\)
Hence, the point \(\mathrm{P}(-3,1)\)
Thus, area of \(\triangle \mathrm{PAB}\) is
\(2 \times \frac{1}{2} \times(1-(-3)) \times(3-1)=8 \text { sq unit. }\)
JEE Main-25.07.2022
Parabola
120247
Let \(\alpha_1\) and \(\alpha_2\) be the ordinates of two points \(A\) and \(B\) on a parabola \(y^2=4 a x\) and let \(\alpha_3\) be the ordinate of the point of intersection of its tangents at \(A\) and \(B\). Then, \(\alpha_3-\alpha_2=\)
1 \(\alpha_3-\alpha_1\)
2 \(\alpha_3+\alpha_1\)
3 \(\alpha_1\)
4 \(\alpha_1-\alpha_3\)
Explanation:
D
Let \(\alpha_1\left(\mathrm{at}_1^2, 2 \mathrm{at}_1\right) \& \alpha_2\left(\mathrm{at}_2^2, 2 \mathrm{at}_2\right)\)
\(\text { Then, } \mathrm{y}_1=2 \mathrm{at}_1 \& \mathrm{y}_2=2 \mathrm{at}_2 \alpha_3 \text { will be }\left(\mathrm{at}_1 \mathrm{t}_2, \mathrm{a}\left(\mathrm{t}_1+\mathrm{t}_2\right)\right)\)
\(\therefore \quad \mathrm{y}_3=\mathrm{a}\left(\mathrm{t}_1+\mathrm{t}_2\right)\)
\(\mathrm{y}_3=\frac{\mathrm{y}_1+\mathrm{y}_2}{2}\)
\(\text { So, } \quad 2 \alpha_3=\alpha_1+\alpha_2\)
\(\text { Or } \quad \alpha_3+\alpha_3=\alpha_1+\alpha_2\)
\(\alpha_3-\alpha_2=\alpha_1-\alpha_3\)
120244
The equations of two sides of a variable triangle are \(x=0\) and \(y=3\), and its third side is a tangent to the parabola \(y^2=6 x\). The locus of its circum centre is:
1 \(4 y^2-18 y+3 x+18=0\)
2 \(4 y^2-18 y-3 x+18=0\)
3 \(4 y^2+18 y+3 x+18=0\)
4 \(4 y^2-18 y-3 x-18=0\)
Explanation:
A Given,
\(y^2=6 x\)
\(y^2=4 \times\left(\frac{3}{2}\right) x\)
\(a=\frac{3}{2}\)
Now, equation of line :
\(\mathrm{y}=\mathrm{mx}+\frac{3}{2 \mathrm{~m}}\)
\(\mathrm{h}=\frac{6 \mathrm{~m}-3}{4 \mathrm{~m}^2}, \mathrm{k}=\frac{6 \mathrm{~m}+3}{4 \mathrm{~m}}\)
Now, eliminating ' \(\mathrm{m}\) ' we get -
\(3 \mathrm{x}=2\left(-2 \mathrm{k}^2+9 \mathrm{k}-9\right)\)
\(4 \mathrm{y}^2-18 \mathrm{y}+3 \mathrm{x}+18=0\)
JEE Main-25.01.2023
Parabola
120245
The parabolas : \(\mathrm{ax}^2+2 \mathrm{bx}+\mathrm{cy}=0\) and \(\mathrm{dx}^2+\) \(2 e x+f y=0\) intersect on the line \(y=1\). If a, b, \(c\), \(d, e, f\) are positive real numbers and \(a, b, c\), are in G.P., then
1 \(\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in G.P
2 d, e, f are in A.P
3 d, e, f are in G.P
4 \(\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in A.P
Explanation:
D Given,
\(\mathrm{P}_1: \mathrm{ax}^2+2 \mathrm{bx}+\mathrm{cy}=0\)
\(\mathrm{P}_2: d \mathrm{dx}^2+2 \mathrm{ex}+\mathrm{fy}=0\)
From \(\mathrm{P}_1: \mathrm{ax}^2+2 \sqrt{\mathrm{ac}} \mathrm{x}+\mathrm{cy}=0 \quad\left\{\because \mathrm{b}^2=\mathrm{ac}\right\}\)
\((x \sqrt{a}+\sqrt{c})^2=0\)
\(x=-\sqrt{\frac{c}{a}}\)
From, \(\mathrm{P}_2: \mathrm{d}\left(\frac{\mathrm{c}}{\mathrm{a}}\right)+2 \mathrm{e}\left(-\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\right)+\mathrm{f}=0\) from \(\left.=(\mathrm{i})\right\}\)
\(\Rightarrow \frac{\mathrm{dc}}{\mathrm{a}}+\mathrm{f}=2 \mathrm{e} \sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\)
\(\Rightarrow \frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{f}}{\mathrm{c}}=2 \mathrm{e} \sqrt{\frac{1}{\mathrm{ac}}}\)
\(\Rightarrow \frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{f}}{\mathrm{c}}=\frac{2 \mathrm{e}}{\mathrm{b}}\)
\(\therefore \quad \frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in AP.
JEE Main-30.01.2023
Parabola
120246
The tangents at the points \(A(1,3)\) and \(B(1,-1)\) on the parabola \(y^2-2 x-2 y=1\) meet at the point \(P\). Then the area (in unit \({ }^2\) ) of the triangle PAB is :
1 4
2 6
3 7
4 8
Explanation:
D The given parabola is
\(y^2-2 x-2 y=1\)
\((y-1)^2=2(x+1)\)
Or
Now, from the above given information the tangent at the point \(A(1,3)\) on the parabola \(y^2-2 x-2 y=1\) is
\(y(3)-(x+1)-(y+3)=1\)
\(3 y-x-1-y-3=1\)
\(2 y-x=5\)
And tangent at \((1,-1)\) is
\(y(-1)-(x+1)-(y-1)=1\)
\(-y-x-1-y+1=1\)
\(x+2 y=-1\)
Now, solving equitation (i) and (ii), we get-
\(x=-3 \text { and } y=1\)
Hence, the point \(\mathrm{P}(-3,1)\)
Thus, area of \(\triangle \mathrm{PAB}\) is
\(2 \times \frac{1}{2} \times(1-(-3)) \times(3-1)=8 \text { sq unit. }\)
JEE Main-25.07.2022
Parabola
120247
Let \(\alpha_1\) and \(\alpha_2\) be the ordinates of two points \(A\) and \(B\) on a parabola \(y^2=4 a x\) and let \(\alpha_3\) be the ordinate of the point of intersection of its tangents at \(A\) and \(B\). Then, \(\alpha_3-\alpha_2=\)
1 \(\alpha_3-\alpha_1\)
2 \(\alpha_3+\alpha_1\)
3 \(\alpha_1\)
4 \(\alpha_1-\alpha_3\)
Explanation:
D
Let \(\alpha_1\left(\mathrm{at}_1^2, 2 \mathrm{at}_1\right) \& \alpha_2\left(\mathrm{at}_2^2, 2 \mathrm{at}_2\right)\)
\(\text { Then, } \mathrm{y}_1=2 \mathrm{at}_1 \& \mathrm{y}_2=2 \mathrm{at}_2 \alpha_3 \text { will be }\left(\mathrm{at}_1 \mathrm{t}_2, \mathrm{a}\left(\mathrm{t}_1+\mathrm{t}_2\right)\right)\)
\(\therefore \quad \mathrm{y}_3=\mathrm{a}\left(\mathrm{t}_1+\mathrm{t}_2\right)\)
\(\mathrm{y}_3=\frac{\mathrm{y}_1+\mathrm{y}_2}{2}\)
\(\text { So, } \quad 2 \alpha_3=\alpha_1+\alpha_2\)
\(\text { Or } \quad \alpha_3+\alpha_3=\alpha_1+\alpha_2\)
\(\alpha_3-\alpha_2=\alpha_1-\alpha_3\)
120244
The equations of two sides of a variable triangle are \(x=0\) and \(y=3\), and its third side is a tangent to the parabola \(y^2=6 x\). The locus of its circum centre is:
1 \(4 y^2-18 y+3 x+18=0\)
2 \(4 y^2-18 y-3 x+18=0\)
3 \(4 y^2+18 y+3 x+18=0\)
4 \(4 y^2-18 y-3 x-18=0\)
Explanation:
A Given,
\(y^2=6 x\)
\(y^2=4 \times\left(\frac{3}{2}\right) x\)
\(a=\frac{3}{2}\)
Now, equation of line :
\(\mathrm{y}=\mathrm{mx}+\frac{3}{2 \mathrm{~m}}\)
\(\mathrm{h}=\frac{6 \mathrm{~m}-3}{4 \mathrm{~m}^2}, \mathrm{k}=\frac{6 \mathrm{~m}+3}{4 \mathrm{~m}}\)
Now, eliminating ' \(\mathrm{m}\) ' we get -
\(3 \mathrm{x}=2\left(-2 \mathrm{k}^2+9 \mathrm{k}-9\right)\)
\(4 \mathrm{y}^2-18 \mathrm{y}+3 \mathrm{x}+18=0\)
JEE Main-25.01.2023
Parabola
120245
The parabolas : \(\mathrm{ax}^2+2 \mathrm{bx}+\mathrm{cy}=0\) and \(\mathrm{dx}^2+\) \(2 e x+f y=0\) intersect on the line \(y=1\). If a, b, \(c\), \(d, e, f\) are positive real numbers and \(a, b, c\), are in G.P., then
1 \(\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in G.P
2 d, e, f are in A.P
3 d, e, f are in G.P
4 \(\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in A.P
Explanation:
D Given,
\(\mathrm{P}_1: \mathrm{ax}^2+2 \mathrm{bx}+\mathrm{cy}=0\)
\(\mathrm{P}_2: d \mathrm{dx}^2+2 \mathrm{ex}+\mathrm{fy}=0\)
From \(\mathrm{P}_1: \mathrm{ax}^2+2 \sqrt{\mathrm{ac}} \mathrm{x}+\mathrm{cy}=0 \quad\left\{\because \mathrm{b}^2=\mathrm{ac}\right\}\)
\((x \sqrt{a}+\sqrt{c})^2=0\)
\(x=-\sqrt{\frac{c}{a}}\)
From, \(\mathrm{P}_2: \mathrm{d}\left(\frac{\mathrm{c}}{\mathrm{a}}\right)+2 \mathrm{e}\left(-\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\right)+\mathrm{f}=0\) from \(\left.=(\mathrm{i})\right\}\)
\(\Rightarrow \frac{\mathrm{dc}}{\mathrm{a}}+\mathrm{f}=2 \mathrm{e} \sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\)
\(\Rightarrow \frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{f}}{\mathrm{c}}=2 \mathrm{e} \sqrt{\frac{1}{\mathrm{ac}}}\)
\(\Rightarrow \frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{f}}{\mathrm{c}}=\frac{2 \mathrm{e}}{\mathrm{b}}\)
\(\therefore \quad \frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in AP.
JEE Main-30.01.2023
Parabola
120246
The tangents at the points \(A(1,3)\) and \(B(1,-1)\) on the parabola \(y^2-2 x-2 y=1\) meet at the point \(P\). Then the area (in unit \({ }^2\) ) of the triangle PAB is :
1 4
2 6
3 7
4 8
Explanation:
D The given parabola is
\(y^2-2 x-2 y=1\)
\((y-1)^2=2(x+1)\)
Or
Now, from the above given information the tangent at the point \(A(1,3)\) on the parabola \(y^2-2 x-2 y=1\) is
\(y(3)-(x+1)-(y+3)=1\)
\(3 y-x-1-y-3=1\)
\(2 y-x=5\)
And tangent at \((1,-1)\) is
\(y(-1)-(x+1)-(y-1)=1\)
\(-y-x-1-y+1=1\)
\(x+2 y=-1\)
Now, solving equitation (i) and (ii), we get-
\(x=-3 \text { and } y=1\)
Hence, the point \(\mathrm{P}(-3,1)\)
Thus, area of \(\triangle \mathrm{PAB}\) is
\(2 \times \frac{1}{2} \times(1-(-3)) \times(3-1)=8 \text { sq unit. }\)
JEE Main-25.07.2022
Parabola
120247
Let \(\alpha_1\) and \(\alpha_2\) be the ordinates of two points \(A\) and \(B\) on a parabola \(y^2=4 a x\) and let \(\alpha_3\) be the ordinate of the point of intersection of its tangents at \(A\) and \(B\). Then, \(\alpha_3-\alpha_2=\)
1 \(\alpha_3-\alpha_1\)
2 \(\alpha_3+\alpha_1\)
3 \(\alpha_1\)
4 \(\alpha_1-\alpha_3\)
Explanation:
D
Let \(\alpha_1\left(\mathrm{at}_1^2, 2 \mathrm{at}_1\right) \& \alpha_2\left(\mathrm{at}_2^2, 2 \mathrm{at}_2\right)\)
\(\text { Then, } \mathrm{y}_1=2 \mathrm{at}_1 \& \mathrm{y}_2=2 \mathrm{at}_2 \alpha_3 \text { will be }\left(\mathrm{at}_1 \mathrm{t}_2, \mathrm{a}\left(\mathrm{t}_1+\mathrm{t}_2\right)\right)\)
\(\therefore \quad \mathrm{y}_3=\mathrm{a}\left(\mathrm{t}_1+\mathrm{t}_2\right)\)
\(\mathrm{y}_3=\frac{\mathrm{y}_1+\mathrm{y}_2}{2}\)
\(\text { So, } \quad 2 \alpha_3=\alpha_1+\alpha_2\)
\(\text { Or } \quad \alpha_3+\alpha_3=\alpha_1+\alpha_2\)
\(\alpha_3-\alpha_2=\alpha_1-\alpha_3\)
120244
The equations of two sides of a variable triangle are \(x=0\) and \(y=3\), and its third side is a tangent to the parabola \(y^2=6 x\). The locus of its circum centre is:
1 \(4 y^2-18 y+3 x+18=0\)
2 \(4 y^2-18 y-3 x+18=0\)
3 \(4 y^2+18 y+3 x+18=0\)
4 \(4 y^2-18 y-3 x-18=0\)
Explanation:
A Given,
\(y^2=6 x\)
\(y^2=4 \times\left(\frac{3}{2}\right) x\)
\(a=\frac{3}{2}\)
Now, equation of line :
\(\mathrm{y}=\mathrm{mx}+\frac{3}{2 \mathrm{~m}}\)
\(\mathrm{h}=\frac{6 \mathrm{~m}-3}{4 \mathrm{~m}^2}, \mathrm{k}=\frac{6 \mathrm{~m}+3}{4 \mathrm{~m}}\)
Now, eliminating ' \(\mathrm{m}\) ' we get -
\(3 \mathrm{x}=2\left(-2 \mathrm{k}^2+9 \mathrm{k}-9\right)\)
\(4 \mathrm{y}^2-18 \mathrm{y}+3 \mathrm{x}+18=0\)
JEE Main-25.01.2023
Parabola
120245
The parabolas : \(\mathrm{ax}^2+2 \mathrm{bx}+\mathrm{cy}=0\) and \(\mathrm{dx}^2+\) \(2 e x+f y=0\) intersect on the line \(y=1\). If a, b, \(c\), \(d, e, f\) are positive real numbers and \(a, b, c\), are in G.P., then
1 \(\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in G.P
2 d, e, f are in A.P
3 d, e, f are in G.P
4 \(\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in A.P
Explanation:
D Given,
\(\mathrm{P}_1: \mathrm{ax}^2+2 \mathrm{bx}+\mathrm{cy}=0\)
\(\mathrm{P}_2: d \mathrm{dx}^2+2 \mathrm{ex}+\mathrm{fy}=0\)
From \(\mathrm{P}_1: \mathrm{ax}^2+2 \sqrt{\mathrm{ac}} \mathrm{x}+\mathrm{cy}=0 \quad\left\{\because \mathrm{b}^2=\mathrm{ac}\right\}\)
\((x \sqrt{a}+\sqrt{c})^2=0\)
\(x=-\sqrt{\frac{c}{a}}\)
From, \(\mathrm{P}_2: \mathrm{d}\left(\frac{\mathrm{c}}{\mathrm{a}}\right)+2 \mathrm{e}\left(-\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\right)+\mathrm{f}=0\) from \(\left.=(\mathrm{i})\right\}\)
\(\Rightarrow \frac{\mathrm{dc}}{\mathrm{a}}+\mathrm{f}=2 \mathrm{e} \sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\)
\(\Rightarrow \frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{f}}{\mathrm{c}}=2 \mathrm{e} \sqrt{\frac{1}{\mathrm{ac}}}\)
\(\Rightarrow \frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{f}}{\mathrm{c}}=\frac{2 \mathrm{e}}{\mathrm{b}}\)
\(\therefore \quad \frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}\) are in AP.
JEE Main-30.01.2023
Parabola
120246
The tangents at the points \(A(1,3)\) and \(B(1,-1)\) on the parabola \(y^2-2 x-2 y=1\) meet at the point \(P\). Then the area (in unit \({ }^2\) ) of the triangle PAB is :
1 4
2 6
3 7
4 8
Explanation:
D The given parabola is
\(y^2-2 x-2 y=1\)
\((y-1)^2=2(x+1)\)
Or
Now, from the above given information the tangent at the point \(A(1,3)\) on the parabola \(y^2-2 x-2 y=1\) is
\(y(3)-(x+1)-(y+3)=1\)
\(3 y-x-1-y-3=1\)
\(2 y-x=5\)
And tangent at \((1,-1)\) is
\(y(-1)-(x+1)-(y-1)=1\)
\(-y-x-1-y+1=1\)
\(x+2 y=-1\)
Now, solving equitation (i) and (ii), we get-
\(x=-3 \text { and } y=1\)
Hence, the point \(\mathrm{P}(-3,1)\)
Thus, area of \(\triangle \mathrm{PAB}\) is
\(2 \times \frac{1}{2} \times(1-(-3)) \times(3-1)=8 \text { sq unit. }\)
JEE Main-25.07.2022
Parabola
120247
Let \(\alpha_1\) and \(\alpha_2\) be the ordinates of two points \(A\) and \(B\) on a parabola \(y^2=4 a x\) and let \(\alpha_3\) be the ordinate of the point of intersection of its tangents at \(A\) and \(B\). Then, \(\alpha_3-\alpha_2=\)
1 \(\alpha_3-\alpha_1\)
2 \(\alpha_3+\alpha_1\)
3 \(\alpha_1\)
4 \(\alpha_1-\alpha_3\)
Explanation:
D
Let \(\alpha_1\left(\mathrm{at}_1^2, 2 \mathrm{at}_1\right) \& \alpha_2\left(\mathrm{at}_2^2, 2 \mathrm{at}_2\right)\)
\(\text { Then, } \mathrm{y}_1=2 \mathrm{at}_1 \& \mathrm{y}_2=2 \mathrm{at}_2 \alpha_3 \text { will be }\left(\mathrm{at}_1 \mathrm{t}_2, \mathrm{a}\left(\mathrm{t}_1+\mathrm{t}_2\right)\right)\)
\(\therefore \quad \mathrm{y}_3=\mathrm{a}\left(\mathrm{t}_1+\mathrm{t}_2\right)\)
\(\mathrm{y}_3=\frac{\mathrm{y}_1+\mathrm{y}_2}{2}\)
\(\text { So, } \quad 2 \alpha_3=\alpha_1+\alpha_2\)
\(\text { Or } \quad \alpha_3+\alpha_3=\alpha_1+\alpha_2\)
\(\alpha_3-\alpha_2=\alpha_1-\alpha_3\)
