What are the secondary axes of a spherical mirror ?

What are the secondary axes of a spherical mirror ?
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- Innumerable
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Last Answer : : The point at which the rays of light parallel to the main axis of the spherical mirror meet at the point above the main axis (in the concave mirror) or appear to be moving away from the point above the main axis (in the convex mirror) is called the main focus of the spherical mirror.

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Last Answer : The center point of the reflecting surface of a spherical mirror is called pole.

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The point at which the two coordinate axes meet is called the -Maths 9th

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Find the equation of the circle which touches the both axes in first quadrant and whose radius is a. -Maths 9th

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Last Answer : Given that the circle of radius ‘a’ touches both axis. So, its centre is (a, a) So, the equation of required circles is : (x-a)2 + (y-a)2 = a2 ⇒ x2 - 2ax + a2+y2 - 2ay + a2 = a2 ⇒ x2 + y2 - 2ax - 2ay + a2 = 0

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What do you mean by Rectangular Axes? -Maths 9th

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Last Answer : (c) \(rac{19}{90}\)In the words ASSISTANT' and STATISTICS', N' and C' are the uncommon letters. The same letters are A, I, S and T whose numbers in both the words are as follows:AISTASSISTANT\( ightarrow\)2132STATISTICS\( ightarrow\ ... 1}{45}\) + \(rac{1}{10}\) + \(rac{1}{15}\) = \(rac{19}{90}\).

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Last Answer : Let the co-ordinates of the point of internal division A be (x, y). Then,\(x\) = \(rac{2 imes(-7)+3 imes8}{2+3}\) = \(rac{-14+24}{5}\) = \(rac{10}{5}\) = 2y = \(rac{2 imes4+3 imes9}{2+3}\) = \(rac{8+27}{5}\) = \(rac{35}{5}\) = 7∴ Co-ordinates of the point for internal division are (2, 7).

What is the equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinates axes whose sum is –1 ? -Maths 9th

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Last Answer : Diagonals of a rhombus bisect each other at right angles ⇒ Co-ordinates of mid-points of AC and BD are equal∴ 0 = \(\bigg(rac{4+(-2)}{2},rac{-5+(-1)}{2}\bigg)\) = (1, -3)Slope of BD = \(rac{-5+1}{4+2}\) = \(rac{-4}{6}\) ... (rac{3}{2}\) isy + 3 = \(rac{3}{2}\) (x - 1)⇒ 2y + 6 = 3x - 3 ⇒ 2y = 3x - 9.

If (–5, 4) divides the line segment between the co-ordinate axes in the ratio 1 : 2, then what is its equation ? -Maths 9th

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Last Answer : (d) x = yThe equations of the given lines are: 4x + 3y = 12 ...(i) 3x + 4y = 12 ...(ii) Solving the simultaneous equations (i) and (ii), we get\(x\) = \(rac{12}{7}\), y = \(rac{12}{7}\)∴ Point of the ... )isy - 0 = \(\bigg(rac{rac{12}{7}-0}{rac{12}{7}-0}\bigg)\) (x - 0), i.e., y = x.

The line through the points (4, 3) and (2, 5) cuts off intercepts of lengths λ and μ on the axes. Which one of the following is correct ? -Maths 9th

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Last Answer : (c) a, b, c are in H.P. only for all m As the points A(a, ma), B[b, (m + 1)b] and C[c, (m + 2)c] are collinear. Area of Δ ABC should be equal to zero.⇒ \(rac{1}{2}\)[x1(y2 - y3) + x2(y3 - y1) + ... - bc = 0 ⇒ ab + bc = 2ac ⇒ b = \(rac{2ac}{a+c}\)∴ a, b, c are harmonic progression (H.P.) for all m.

A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

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Last Answer : (d) x + 2y = 2Let the required equation make intercept on x-axis = 2a ⇒ intercept made on y-axis = a ∴ Eqn of the given line in the intercept from:\(rac{x}{2a}+rac{y}{a}=1\) ...(i)Since the line ... 1 ⇒ a = 1.∴ Required equation of line : \(rac{x}{2 imes1}+rac{y}{1}=1\) ⇒ x + 2y = 2.

A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of -Maths 9th

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Last Answer : (d) \(rac{23}{\sqrt{17}}\)The given lines are:L : \(rac{x}{5}+rac{y}{b}=1\) ....(i)K : \(rac{x}{c}+rac{y}{3}=1.\) ...(ii)Since line L passes through (13, 32),\(rac{13}{ ... between parallel lines ax + by + c1 = 0 and ax + by c2 = 0 is d = \(rac{|c_2-c_1|}{\sqrt{a^2+b^2}}\bigg)\)

The straight line ax + by + c = 0 and the co-ordinate axes form an isosceles triangle under which of the following conditions ? -Maths 9th

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Last Answer : (a) | a | = | b | The equation of line AB, i.e., ax + by + c = 0 in intercept form is ax + by = - c⇒ \(rac{x}{\big(-rac{c}{a}\big)}\) + \(rac{x}{\big(-rac{c}{b}\big)}\) = 1Δ AOB is isosceles Δ if OA = OB, i.e., ... \(rac{-c}{a}\) = \(rac{-c}{a}\) ⇒ \(rac{1}{a}\) = \(rac{1}{a}\) ⇒ | a | = | b |.

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