1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

If the parabolas y2 = 4b(x – c) and y2 = 8ax have a common normal, then which on of the following is a valid choice for the ordered triad (a, b, c) ?
A
(1, 1, 3)
B
(1, 1, 0)
C
$$\left( {{1 \over 2},2,0} \right)$$
D
$$\left( {{1 \over 2},2,3} \right)$$

Explanation

Normal to the two given curves are

y = m(x – c) – 2bm – bm3,

y = mx – 4am – 2am3

If they have a common normal, then

(c + 2b)m + bm3 = 4am + 2am3

$$ \Rightarrow $$ (4a – c – 2b) m = (b – 2a)m3

$$ \Rightarrow $$ (4a – c – 2b) = (b – 2a)m2

$$ \Rightarrow $$ m2 = $${c \over {2a - b}} - 2$$ $$>$$ 0

By checking all options we found (A) is right option.

Note : We get that all the options are correct for m = 0 i.e., when common normal is x-axis. But may be in question they want common normal other than x – axis, hence answer is (A).
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

The equation of a tangent to the hyperbola 4x2 – 5y2 = 20 parallel to the line x – y = 2 is -
A
x $$-$$ y + 9 = 0
B
x $$-$$ y $$-$$ 3 = 0
C
x $$-$$ y + 1 = 0
D
x $$-$$ y + 7 = 0

Explanation

Hyperbola $${{{x^2}} \over 5} - {{{y^2}} \over 4} = 1$$

slope of tangent = 1

equation of tangent y = x $$ \pm $$ $$\sqrt {5 - 4} $$

$$ \Rightarrow $$  y = x $$ \pm $$ 1

$$ \Rightarrow $$  y = x + 1   

or

$$ \Rightarrow $$  y = x $$-$$ 1
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

The length of the chord of the parabola x2 $$=$$ 4y having equation x – $$\sqrt 2 y + 4\sqrt 2 = 0$$  is -
A
$$8\sqrt 2 $$
B
$$6\sqrt 3 $$
C
$$3\sqrt 2 $$
D
$$2\sqrt {11} $$

Explanation

x2 = 4y

x $$-$$ $$\sqrt 2 $$y + 4$$\sqrt 2 $$ = 0

Solving together we get

x2 = 4$$\left( {{{x + 4\sqrt 2 } \over {\sqrt 2 }}} \right)$$

$$\sqrt 2 $$x2 + 4x + 16$$\sqrt 2 $$

$$\sqrt 2 $$x2 $$-$$ 4x $$-$$ 16$$\sqrt 2 $$ = 0

x1 + x2 = 2$$\sqrt 2 $$; x1x2 = $${{ - 16\sqrt 2 } \over {\sqrt 2 }}$$ = $$-$$ 16

Similarly,

($$\sqrt 2 $$y $$-$$ 4$$\sqrt 2 $$)2 = 4y

2y2 + 32 $$-$$ 16y = 4y



$$\ell $$AB = $$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$

$$ = \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)} $$

$$ = \sqrt {8 + 64 + 100 - 64} $$

$$ = \sqrt {108} = 6\sqrt 3 $$
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

Let S = $$\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r \ne \pm 1.$$ Then S represents
A
an ellipse whose eccentricity is $${1 \over {\sqrt {r + 1} }},$$ where r > 1
B
an ellipse whose eccentricity is $${2 \over {\sqrt {r + 1} }},$$ where 0 < r < 1
C
an ellipse whose eccentricity is $${2 \over {\sqrt {r - 1} }},$$ where 0 < r < 1
D
an ellipse whose eccentricity is $${2 \over {\sqrt {r + 1} }},$$ where r > 1

Explanation

$${{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}} = 1$$

for r > 1,     $${{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1$$

$$e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)} $$

$$ = \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}} $$

$$ = \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}} $$

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