True False[1 Marks ]

Question 11 Mark

The point (1, 2) lies inside the circle x² + y² - 2x + 6y + 1 = 0.

Answer

False.
Solution:
Given equation of circle is x² + y² - 2x + 6y + 1 = 0
Here 2g = -2 ⇒ g = -1
2f = 6 ⇒ f = 3
$\therefore$ Centre = (-g, -f) = (1, -3)
and $\text{r}=\sqrt{\text{g}^2+\text{f}^2-\text{c}}=\sqrt{1+9-1}=3$
$\therefore$ Distance between the point lies outside the circle.
Hence, the given statement is False.

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Question 21 Mark

If P is a point on the ellipse $\frac{\text{x}^2}{16}+\frac{\text{y}^2}{25}=1$ whose foci are S and S', then PS + PS' = 8.

Answer

False.
Solution:
We have equation of the qllipse is $\frac{\text{x}^2}{16}+\frac{\text{y}^2}{25}=1$
From the definition of the ellipse, we know that sum of the distance of any point P on the ellipse from the foci is equal to the length of the major axis.
Here major axis = 2b = 2 × 5 = 10
S and S' are foci, then SP + S'P = 10

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Question 31 Mark

The line 2x + 3y = 12 touches the ellipse $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=2$ at the point (3, 2).

Answer

True.
Solution:
If line 2x + 3y = 12 touches the ellipse $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=2,$ at the point (3, 2) satisfies both line and ellipse.
$\therefore$ For line 2x + 3y = 12
2(3) + 3(2) = 12
6 + 6 = 12
12 = 12 True.
For ellipse $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=2$
$\frac{(3)^2}{9}+\frac{(2)^2}{4}=2$
$\frac{9}{9}+\frac{4}{4}=2$
$1+1=2$
$2=2\text{ True}$
Hence, the given statement is true.

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Question 41 Mark

The shortest distance from the point (2, -7) to the circle x² + y² - 14x - 10y - 151 = 0 is equal to 5.
[Hint: The shortest distance is equal to the difference of the radius and the distance between the centre and the given point]

Answer

False.
Solution:
Given circle is x² + y² - 14x - 10y - 151 = 0
$\therefore\text{ Centre}\equiv\text{C}(7,5)$
And $\text{Radius}=\sqrt{49+25+151}=\sqrt{225}=15$
Now distance between the point P(2, -7) and centre,
$=\sqrt{(2-7)^2+(-7-5)^2}=\sqrt{25+144}=\sqrt{169}=13$
$\therefore$ Shortest distance of point P from the circle = |13 - 15| = 2

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Question 51 Mark

The locus of the point of intersection of lines $\sqrt{3}\text{x}-\text{y}-4\sqrt{3}\text{k}=0$ and $\sqrt{3}\text{kx}+\text{ky}-4\sqrt{3}=0$ for different value of k is a hyperbola whose eccentricity is 2.
[Hint: Eliminate k between the given equations]

Answer

True.
Solution:
Given equation of lines are,
$\sqrt{3}\text{x}-\text{y}-4\sqrt{3}\text{k}=0\ ...(\text{i})$
and $\sqrt{3}\text{kx}+\text{ky}-4\sqrt{3}=0\ ....(\text{ii})$
From Eq. (i), $\text{k}=\frac{\sqrt{3}\text{x}-\text{y}}{4\sqrt{3}}$
From Eq. (ii), $\text{k}=\frac{4\sqrt{3}}{\sqrt{3}\text{x}+\text{y}}$
Equating the values of k, we get
$\frac{\sqrt{3}\text{x}-\text{y}}{4\sqrt{3}}=\frac{4\sqrt{3}}{\sqrt{3}\text{x}+\text{y}}$
$\Rightarrow3\text{x}^2-\text{y}^2=48$
$\Rightarrow\frac{\text{x}^2}{16}-\frac{\text{y}^2}{48}=1,$ which is equation of hyperbola
$\therefore\ \text{a}^2=16$ and $\text{b}^2=48$
$\Rightarrow\text{e}^2=1+\frac{48}{16}=1+3=4$
$\Rightarrow\text{e}=2$

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Question 61 Mark

The line x + 3y = 0 is a diameter of the circle x² + y² + 6x + 2y = 0.

Answer

False.
Solution:
Given equation of the circle is,
x² + y² + 6x + 2y = 0
Centre is (-3, -1)
If x + 3y = 0 is the equation of diameter, then the centre (-3, -1) will lie on x + 3y = 0
-3 + 3(-1) = 0
⇒ -6 ≠ 0
So, x + 3y = 0 is not the diameter of the circle.
Hence, he given statement is False.

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Question 71 Mark

The line lx + my + n = 0 will touch the parabola y² = 4ax if ln = am²

Answer

True.
Solution:
Give line lx + my + n = 0 and parabola y² = 4ax
Solving line and parabola for their point of intersection, we get
$\frac{1}{4\text{a}}\text{y}^2+\text{my}+\text{n}=0$
Since line touches the parabola, above equation must have equal roots.
$\therefore\ \text{Discriminant},\text{ D}=0$
$\therefore\text{ m}^2-4\Big(\frac{1}{4\text{a}}\Big)\text{n}=0$
$\Rightarrow\text{ am}^2=\text{nl}$

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Question 81 Mark

If the line lx + my = 1 is a tangent to the circle x² + y² = a², then the point (l, m) lies on a circle.
[Hint: Use that distance from the centre of the circle to the given line is equal to radius of the circle]

Answer

True.
Solution:
Given equation of circle is x² + y² = a²
and the tangent is lx + my = 1
Here centre is (0, 0) and radius = a
If (l, m) lie on the circle
$\therefore\ \sqrt{(1-0)^2+(\text{m}-0)^2}=\text{a}$
$\Rightarrow\sqrt{\text{l}^2+\text{m}^2}=\text{a}$
$\Rightarrow\text{l}^2+\text{m}^2=\text{a}^2$ (which is a circle)
So, the point (l, m) lies on the circle.
Hence, the given statement is True.

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MATHS True False[1 Marks ]

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